Typesetting math: 69%
BYJU'S Exercise 13.1
Q1.
Evaluate limx→3x+5 limit:
Soln:
limx→3x+5=3+5=8
Q2.
Evaluate limx→π(x–227)
Soln:
limx→π(x–227)=(π–227)
Q3.
Evaluate limr→lπr2
Soln:
limr→lπr2=π(l)2=π
Q4.
Evaluate limx+44x+6x–3
Soln:
limx+44x+6x–3=4(4)+64–3=16+61=221=22
Q5.
Evaluate limx→−1x12+x7+1x–1
Soln:
limx→−1x12+x7+1x–1=(−1)12+(−1)7+1−1–1=1–1+1−2=−12
Q6.
Evaluate limx→0(x+3)5–3x
Soln:
limx→0(x+3)5–3x
Put x + 3 = y so that y→1asx→0
Accordingly, limx→0(x+3)5–3x=limy→1y5–3y–3=limy→1y5–35y–3=5.35–1=405∴limx→0(x+3)5–3x=405
Q7.
Evaluate limx→23x2–x–10x2–4
Soln:
At x = 2, the given rational function has the value of 0/ 0
limx→23x2–x–10x2–4=limx→2(x–2)(3x+5)(x–2)(x+2)=limx→23x+5x+2=3(2)+52+2=114
Q8.
Evaluate limx→3x4–812x2–5x–3
Soln:
At x = 2, the given rational function has the value of 0/ 0
limx→3x4–812x2–5x–3=limx→3(x−3)(x+3)(x2+9)(x–3)(2x+1)=limx→3(x+3)(x2+9)2x+1=(3+3)(32+9)2(3)+1=6×187=1087
Q9.
Evaluate limx→0ax+bcx+1
Soln:
limx→0ax+bcx+1=a(0)+bc(0)+1=b
Q10.
Evaluate limz→1z13–1z16–1
Soln:
limz→1z13–1z16–1
At z = 1, the given function has the value of 0/ 0
Put z16=xsothatz→1asx→1Accordingly,limz→1z13–1z16–1=limx→1x2–1x–1=limx→1x2–12x–1=2.12–1=2∴limz→1z13–1z16–1=2
Q11.
Evaluate limx→1ax2+bx+ccx2+bx+a,a+b+c≠0
Soln:
limx→1ax2+bx+ccx2+bx+a=a(1)2+b(1)+cc(1)2+b(1)+a=a+b+ca+b+c=1
Q12.
limx→21x+12x+2
Soln:
limx→21x+12x+2
At x = -2, the given function has the value of 0/ 0
Now,limx→21x+12x+2=limx→−2(2+x2x)x+2=limx→−212x=12(−2)=−14
Q13.
Evaluate limx→0sinaxbx
Soln:
limx→0sinaxbx
At x = 0, the given function has the value of 0/ 0
Now,limx→0sinaxbx=limx→0sinaxax×axbx=limx→0(sinaxax)×(ab)=ablimax→0(sinaxax)=ab×1=ab
Q14.
Evaluate limx→0sinaxsinbx,a,b≠0
Soln:
limx→0sinaxsinbx,a,b≠0
At x = 0, the given function has the value of 0/ 0
Now,limx→0sinaxsinbx=limx→0(sinaxax)×ax(sinbxbx)×bx=(ab)×limax→0(sinaxax)limbx→0(sinbxbx)=(ab)×11=ab
Q15.
Evaluate limx→πsin(π–x)π(π–x)
Soln:
limx→πsin(π–x)π(π–x)
It seems x→π⇒(π–x)→0 ∴limx→πsin(π–x)π(π–x)=1πlim(z–x)–1sin(π–x)(π–x)=1π×1=1π
Q16.
Evaluate limx→0cosxπ–x
Soln;
limx→0cosxπ–x=cos0π–0=1π
Q17.
Evaluate limx→0cos2x–1cosx–1
Soln:
limx→0cos2x–1cosx–1
At x = 0, the given function has the value of 0/ 0
Now,limx→0cos2x–1cosx–1=limx→01–2sin2x–11–2sin2x2–1=limx→0sin2xsin2x2=limx→0(sin2xx2)×x2⎛⎝sin2x2(x2)2⎞⎠×x24=4limx→0(sin2xx2)limx→0⎛⎝sin2x2(x2)2⎞⎠ =4(limx→0sinxx)2(limx2→0sinx2x2)2=41212=4
Q18.
Evaluate limx→0ax+xcosxbsinx
Soln:
limx→0ax+xcosxbsinx
At x = 0, the given function has the value of 0/ 0
Nowlimx→0ax+xcosxbsinx=1blimx→0x(a+cosx)sinx=1blimx→0(xsinx)×limx→0(a+cosx)=1b×1(limx→0sinxx)×limx→0(a+cosx)=1b×(a+cos0)=a+1b
Q19.
Evaluate limx→0xsecx
Soln:
limx→0xsecx=limx→0xcosx=0cos0=01=0
Q20.
Evaluate limx→0sinax+bxax+sinbxa,ba+b≠0
Soln:
At x =0, the given function has the value of 0/ 0
Now,limx→0sinax+bxax+sinbxlimx→0(sinaxax)ax+bxax+bx(sinbxbx)= =(limax→0sinaxax)×limx→0(ax)+limx→0bxlimx→0ax+limx→0bx(limbx→0sinbxbx)=limx→0(ax)+limx→0bxlimx→0ax+limx→0bx=limx→0(ax+bx)limx→0(ax+bx)=limx→0(1)=1
Q21.
Evaluate limx→0(cscx–cotx)
Soln:
At x =0, the given function has the value of ∞ – ∞
Now,limx→0(cscx–cotx)=limx→0(1sinx–cosxsinx)=limx→0(1–cosxsinx)=limx→0(1–cosxsinx)(sinxx)=limx→01–cosxxlimx→0sinxx=01=0
Q22.
limx→πxtan2xx–π2
Soln:
limx→πxtan2xx–π2
At x=π2, the given function has the value of 0/ 0
Now,putx–π2=ysothatx→π2,y→0∴limx→π2tan2xx–π2=limy→0tan2(y+π2)y=limy→0tan(π+2y)y=limy→0tan2yy=limy→0sin2yycos2y=limy→0(sin2y2y×2cos2y)=(lim2y→0sin2y2y)×limy→0(2cos2y)=1×2cos0=1×21=2
Q23.
Find limx→0f(x)andlimx→0f(x),wheref(x)={2x+33(x+1)x≤0x>0
Soln:
Given function f(x)={2x+33(x+1)x≤0x>0
limx→0f(x)=limx→0[2x+3]=2(0)+3=3
limx→0f(x)=limx→03(x+1)=3(0+1)=3
∴limx→0f(x)=limx→0f(x)=limx→0f(x)=3
limx→1f(x)=limx→13(x+1)=3(1+1)=6
limx→1f(x)=limx→13(x+1)=3(1+1)=6
∴limx→1f(x)=limx→1f(x)=limx→1f(x) = 6
Q24.
Find limx→1f(x),wheref(x)={x2–1,−x2–1,x≤1x≥1
Soln:
Given function is
f(x)={x2–1,−x2–1,x≤1x≥1 limx→1−f(x)=limx→1[x2–1]=12–1=1–1=0 limx→1+f(x)=limx→1[−x2–1]=−12–1=1–1=0
We observed that limx→1−f(x)≠limx→1+f(x)
Hence, limx→1f(x) doesn’t exist.
25.
Evaluate limx→0f(x),wheref(x)={|x|x,0,x≠0x=0
Soln:
Given function f(x)={|x|x,0,x≠0x=0
When |x|=−x limx→0−f(x)=limx→0−[|x|x]
= limx→0(−xx)
= limx→0(−1)
= -1
When |x|=x limx→0+f(x)=limx→0+[|x|x]
= limx→0[−xx]
= limx→0(1)
= 1
We observe that limx→0−f(x)≠limx→0+f(x)
Hence, limx→0f(x) doesn’t exist.
Q26.
Find limx→0f(x),wheref(x)={x|x|,0,x≠0x=0
Soln:
f(x) = {x|x|,0,x≠0x=0
When |x|=−x limx→0−f(x)=limx→0−[x|x|] =limx→0[x–x] =limx→0(–1)
= -1
When |x|=x limx→0+f(x)=limx→0+[x|x|] =limx→0[xx] =limx→0(1)
= 1
We observe that limx→0−f(x)≠limx→0+f(x)
Hence, limx→0f(x) doesn’t exist.
Q27.
Find limx→5f(x),wheref(x)=|x|–5
Soln:
Given function is f(x)=|x|–5
When x>0,|x|=x limx→5−f(x)=limx→5−[|x|–5]
= limx→5(x–5)
= 5 – 5
= 0
When x>0,|x|=x limx→5+f(x)=limx→5+(|x|–5)
= limx→5(x–5)
= 5 – 5
= 0
∴limx→5−f(x)=limx→5+f(x)=0
Hence, limx→5f(x)=0
Q28.
Suppose f(x)=⎧⎩⎨⎪⎪a+bx,4,b–ax,ifx<1ifx=0,ifx>1
And limx→1f(x)=f(1) what can be the values of b and a?
Soln:
Given function
f(x)=⎧⎩⎨⎪⎪a+bx,4,b–ax,ifx<1ifx=0,ifx>1 limx→1−f(x)=limx→1(a+bx)=a+b limx→1−f(x)=limx→1(b–ax)=b–a
f(1) = 4
Given that limx→1f(x)=f(1).
∴limx→1−f(x)=limx→1+f(x)=limx→1f(x)=f(1)
=> a+ b = 4 and b – a = 4
Solving both the equations, we get a = 0 and b = 4.
Hence, b and a are 4 and 0
Q29.
A function f(x)=(x–a1)(x–a2)....(x–an) is define by fixed real numbers a1, a2, . . . . an.
Find limx→a1f(x). For some a≠a1,a2,....,ancomputelimx→anf(x)
Soln:
Given function = f(x)=(x–a1)(x–a2)...(x–an) limx→a1f(x)=limx→a1[(x–a1)(x–a2)...(x–an)]
=[limx→a1(x–a1)][limx→a1(x–a2)]...[limx→a1(x–an)]
=(a1–a1)(a1–a2)....(a1–an)=0
∴limx→a1]f(x)=0
Now, limx→a1]f(x)=limx→a[(x–a1)(x–a2)...(x–an)]
= =limx→a[x–a1][x–a2]...[x–an]
(a1–a1)(a1–a2)....(a1–an) ∴limx→af(x)=(a1–a1)(a1–a2)....(a1–an)
Q30.
If f(x)=⎧⎩⎨⎪⎪|x|+1,0,|x|–1,x≤0x=0x>0
limx→af(x) will exist for what values?
Soln:
Given function:
f(x)=⎧⎩⎨⎪⎪|x|+1,0,|x|–1,x≤0x=0x>0
When a = 0
limx→0−f(x)=limx→0−(|x|+1) =limx→0(–x+1) [if x < 0, |x| = -x]
= -0 + 1
= 1
limx→0+f(x)=limx→0+(|x|–1) =limx→0(x–1) [if x > 0, |x| = x]
= 0 – 1
= -1
We have, limx→0−f(x)≠limx→0+f(x) ∴limx→0f(x) doesn’t exist.
When a < 0
limx→a−f(x)=limx→a−(|x|+1) =limx→a(−x+1)[x<a<0⇒|x|=–x]
= -a + 1
limx→a+f(x)=limx→a+(|x|+1) =limx→a(−x+1)[x<a<0⇒|x|=–x]
= – a + 1
∴limx→a−=limx→a+=–a+1
Hence, at x = a limit of f(x) exist, where a > 0
Thus, limx→af(x)existforalla≠0
Q31.
Evaluate limx→1f(x)
If f(x) satisfies, limx→1f(x)–2x2–1=π
Soln:
limx→1f(x)–2x2–1=π ⇒limx→1(f(x)–2)limx→1(x2–1)=π ⇒limx→1(f(x)–2)=πlimx→1(x2–1) ⇒limx→1(f(x)–2)=π(12–1) ⇒limx→1(f(x)–2)=0 ⇒limx→1f(x)–limx→12=0 ⇒limx→1f(x)–2=0 ∴limx→1f(x)=2
Q32.
If f(x)=⎧⎩⎨⎪⎪ax2+b,bx+a,bx3+ax<00≤x≤1x>1
For what values of a and b does limx→0f(x)andlimx→1f(x) exist?
Soln:
Given function
f(x)=⎧⎩⎨⎪⎪ax2+b,bx+a,bx3+ax<00≤x≤1x>1 limx→0−f(x)=limx→0(ax2+b)
= a(0)2 + b
= b
limx→0+f(x)=limx→0(bx+a)
= b(1) + a
= a + b
∴limx→1−f(x)=limx→1+f(x)=limx→1f(x)
Thus, for integral values of n and m limx→1f(x) exist.
Exercise-13.2
Q1. Find derivative for x2 – 2 at x = 10
Soln:
Let f(x) = x2 – 2
Accordingly,
f′(10)=limh→0f(10+h)–f(10)h
=limh→0[(10+h)2–2]–(102–2)h
=limh→0102+2.10.h+h2–2–102+2h
=limh→020h+h2h
=limh→0(20+h)=(20+0)=0
Hence, derivative for x2 – 2 at x = 10 is 20
Q2. Find derivative
99x at x = 100
Soln:
Let f(x) = 99x,
Accordingly,
f′(100)=limh→0f(100+h)–f(100)h
=limh→099(100+h)–99(100)h
=limh→099×100+99h–99×100h
=limh→099hh
=limh→0(99h)=99
Hence, derivative for 99x at x = 100 is 99
Q3. Find derivative
x at x = 1
Soln:
Let f(x) = x
Accordingly,
f′(1)=limh→0f(1+h)–f(1)h
=limh→0(1+h)–1h
=limh→0hh
=limh→0(1)
= 1
Hence, derivative for x at x = 1 is 1
Q4. Using first principle find derivative
(i) x3 – 27
(ii) (x – 1) (x – 2)
(iii) 1x2
(iv) x+1x–1
Soln:
(i) Let f(x) = x3 – 27
From first principle,
f′(x)=limh→0f(x+h)–f(x)h
=limh→0[(x+h)3–27]–(x3–27)h
=limh→0x3+h3+3x2h+3xh2–x3h
=limh→0h3+3x2h+3xh2h
=limh→0(h2+3x2+3xh)
= 0 + 3x2 + 0 = 3x2
(ii) Let f(x) = (x – 1)(x – 2)
From first principle,
f′(x)=limh→0f(x+h)–f(x)h
=limh→0(x+h–1)(x+h–2)–(x–1)(x–2)h
=limh→0(x2+hx–2x+hx+h2–2h–x–h+2)–(x2–2x–x+2)h
=limh→0(hx+hx+h2–2h−h)h
=limh→0(2hx+h2–3h)h
=limh→0(2x+h–3)
(2x+h–3)
= 2x – 3
(iii) Let f(x)=1x2
From first principle
f′(x)=limh→0f(x+h)–f(x)h
=limh→01(x+h)2–1x2h
=limh→01h[x2–(x+h)2x2(x+h)2]
=limh→01h[x2–x2–h2–2hxx2(x+h)2]
=limh→01h[–h2–2hxx2(x+h)2]
=limh→0[–h–2hxx2(x+h)2]
0–2xx2(x+0)2=−2x3
(iv) Let f(x)=x+1x–1
From first principle,
=limh→0(x+h+1x+h–1–x+1x–1)h
=limh→01h[(x–1)(x+h+1)–(x+1)(x+h–1)(x–1)(x+h–1)]
=limh→01h[(x2+hx+x–x–h–1)–(x2+hx–x+x+h–1)(x−1)(x+h+1)]
=limh→01h[−2h(x–1)(x+h–1)]
=limh→0[−2(x–1)(x+h–1)]
=−2(x–1)(x–1)=−2(x–1)2
Q5. Prove f′(1)=100f′(0)
For the function f(x)=x100100+x9999+…+x22+x+1
Soln:
Given function
f(x)=x100100+x9999+…+x22+x+1
ddxf(x)=ddx[x100100+x9999+…+x22+x+1]
ddxf(x)=ddx(x100100)+ddx(x9999)+…+ddx(x22)+ddx(x)+ddx(1)
Using theorem ddx(xn)=nxn–1, we get
ddxf(x)=100x99100+99x9899+....+2x2+1+0
=x99+x98+….+x+1
∴ ddxf(x)=x99+x98+….+x+1
At x = 0
f'(0) = 1
At x = 1,
f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100
Hence, f’(1) = 100 x f1 (0)
Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an
Soln:
Let f(x)=xn+axn–1+a2xn–2+...+an–1x+an
f′(x)=ddx(xn+axn–1+a2xn–2+...+an–1x+an)
=ddx(xn)+addx(xn–1)+a2ddx(xn–2)+...+an–1ddx(x)+anddx(1)
Using ddxxn=nxn−1, we have
f′(x)=nxn–1+a(n–1)xn–2+a2(n–2)xn−3+…+an–1+an(0)
=nxn–1+a(n–1)xn–2+a2(n–2)xn−3+…+an–1
Q7. Find the derivative for
(i) (x – m)(x – n)
(ii) (ax2 + b)2
(iii) x–ax–b
Soln:
(i) Let f(x) = (x – m)(x – n)
⇒f(x)=x2–(m+n)x+mn
f′(x)=ddx(x2–(m+n)x+mn)
=ddx(x2)–(m+n)ddx(x)+ddx(mn)
Using =ddx(xn)=nxn–1, we get
f′(x)=2x−(m+n)+0=2x–m–n
(ii) Let f(x) = (ax2 + b)2
⇒f(x)=a2x2+2abx+b2
f′(x)=ddx(a2x4+2abx2+b2)=a2ddx(x4)+2abddx(x2)+ddx(b2)
Using ddxxn=nxn–1, we have
f′(x)=a2(4x3)+2ab(2x)+b2(0)
= 4a2x3 + 4abx
= 4ax(ax2 + b)
(iii) Let f(x)=(x–a)(x–b)
f′(x)=ddx(x–ax–b)
Quotient rule,
f′(x)=(x–b)ddx(x–a)–(x–a)ddx(x−b)(x–b)2
=(x–b)(1)–(x–a)(1)(x–b)2
=x–b–x+a(x–b)2
=a–b(x–b)2
Q8. If a is constant find derivative of xn–anx–a
Soln:
Let xn–anx–a
⇒f′(x)=ddx(xn–anx–a)
By Question rule
⇒f′(x)=(x–a)ddx(xn–an)–(xn–an)ddx(x–a)(x–a)2
=(x−a)(nxn–1–0)–(xn–an)(x–a)2
=nxn–anxn–1–xn+an(x–a)2
Q9.
(i) 2x – \frac {3}{4}
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x-3 (5 + 3x)
(iv) x5 (3 – 6x-9)
(v) x-4 (3 – 4x-5)
(vi) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}
Soln.
(i) Let f(x) = 2x – \frac {3}{4}
f'(x) = \frac{d}{dx}\left ( 2x – \frac{3}{4} \right )
= 2\frac{d}{dx}\left ( x \right ) – \frac {d}{dx} \left ( \frac{3}{4} \right )
= 2 – 0
= 2
(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)
By Leibnitz product rule,
f'(x) = (5x^{3} + 3x – 1) \frac{d}{dx}(x – 1) + (x – 1) \frac{d}{dx}(5x^{3} + 3x – 1 )
= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)
= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)
= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3
= 20x3 – 15x2 + 6x – 4
(iii) Let f(x) = x-3 (5 + 3x)
f'(x) = x^{-3}\frac{d}{dx}(5 + 3x) + (5 + 3x)\frac{d}{dx}(x^{-3})
= x^{-3}(0 + 3) + (5 + 3x) (-3x^{-3 – 1} )
= x^{-3}( 3) + (5 + 3x) (-3x^{-4} )
= 3x-3 – 15x-4 – 9x-3
= -6x-3 – 15x-4
= -3 x ^{-3} \left ( 2 + \frac{5}{x} \right )
= \frac{-3 x ^{-3} }{x}\left ( 2x + 5 \right )
= \frac{-3 }{ x ^{4} }\left ( 2x + 5 \right )
(iv) let f(x) = x5 (3 – 6x-9)
From Leibnitz product rule,
f'(x) = x^{5} \frac{d}{dx}(3 – 6x^{-9}) + (3 – 6x^{-9}) \frac{d}{dx}(x^{5})
= x^{5} \left \{ 0 – 6 (-9)x^{-9 – 1} \right \} + (3 – 6x^{-9})(5x^{4})
= x^{5} (54x ^{-10}) + 15x^{4} – 30x^{-5}
= 54x-5 + 15x4 – 30x-5
= 24x-5 + 15x4
= = 15x^{4} + \frac{24}{x^{5}}
(v) Let f(x) = x-4 (3 – 4x-5)
From Leibnitz product rule,
f'(x) = x^{-4}\frac{d}{dx}\left ( 3 – 4x^{-5} \right ) + (3 – 4 x^{-5})\frac{d}{dx}(x^{-4})
= x^{-4}\left \{ 0 – 4 (-5)x^{-5 – 1} + (3 – 4x^{-5})(-4)x^{-4 -1} \right \}
= x-4 (20x-6) + (3 – 4x-5)(-4x-5)
= 36x-10 – 12x-5
= -\frac{12}{x^{5}} + \frac{36}{x^{10}}
(iv) Let f(x) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1} f'(x) = \frac{d}{dx}\left ( \frac{2}{x + 1} \right ) – \frac{d}{dx} \left ( \frac{x^{2}}{3x – 1} \right )
From quotient rule,
f'(x) = \left [ \frac{\left ( x + 1 \right ) \frac{d}{dx}(2) – 2 \frac{d}{dx} (x + 1) }{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1) \frac{d}{dx} (x^{2}) – x^{2} \frac{d}{dx}\left ( 3x – 1 \right )}{(3x – 1)^{2}} \right ]
= \left [ \frac{(x + 1)(0) – 2(1)}{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1)(2x) – (x^{2})(3)}{(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \left [ \frac{6x^{2} – 2x – 3x^{2}}{(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \left [ \frac{3x^{2} – 2x} {(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \frac{x(3x – 2)} {(3x – 1)^{2}}
Q10. Using first principle find derivative of \cos x
Soln:
Let f(x) = \cos x.
According to first principle
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{\cos (x + h) – \cos x}{h}
= \lim_{h \rightarrow 0} \left [\frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \right ]
= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h) – \sin x \sin h }{h} \right ]
= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h)}{h} – \frac{\sin x \sin h }{h} \right ]
=- \cos x \left ( \lim_{h \rightarrow 0} \frac{1 – \cosh}{h} \right ) – \sin x\lim_{h \rightarrow 0}\left ( \frac{sin h}{h} \right )
= = – \cos x (0) – \sin x (1) \left [ \lim_{h \rightarrow 0} \frac{1 – cos h}{h} = 0 \; and \; \lim_{h \rightarrow 0}\frac{\sin h }{h} = 1 \right ]
= – sin x
f’(x) = – sin x
Q11.
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7
(vii) 2tan x – 7sec x
Soln:
(i) Let f’(x) = sin x cos x
According to first principle,
f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) -f(x)}{h}
= \lim_{h \rightarrow 0}\frac{\sin (x + h) \cos(x + h) – \sin x \cos x}{h}
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2\sin (x + h) \cos (x + h) – 2 \sin x cos x \right ]
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ \sin 2(x + h) – \sin 2x \right ]
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2 \cos \frac{2x + 2h + 2x}{2} . \sin \frac{2x + 2h – 2x}{2} \right ]
= \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos \frac{ 4x + 2h }{2} \sin \frac{ 2h }{2} \right ]
\lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (2x + h ) \sin h \right ]
\lim_{h \rightarrow 0} \cos (2x + h ) \lim_{h \rightarrow 0} \frac{\sin h}{h}
= cos(2x + 0). 1
= cos 2x
(ii) Let f(x) = sec x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{\sec (x + h) – \sec x}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos(x + h)}{\cos x \cos(x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \right ) \sin \left ( \frac{x – x – h}{2} \right )}{\cos (x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{2x + h}{2} \right ) \sin \left ( \frac{- h}{2} \right )}{\cos (x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\left [ \sin \left ( \frac{2x + h}{2} \right ) \frac{\sin \left ( \frac{h}{2}\right )}{\left (\frac{h}{2}\right )} \right ]}{\cos (x + h)}
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}. \lim_{h \rightarrow 0}\frac{\sin\left ( \frac{2x + h}{2} \right ) }{\cos (x + h)}
= \frac{1}{\cos x} . 1 . \frac{\sin x}{ \cos x}
= \sec x \tan x
(iii)
Let f(x) = 5 sec x + 4 cos x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
\lim_{h \rightarrow 0}\frac{5\sec (x + h) + 4 \cos(x + h) – \left [ 5 \ sec x + 4 \ cos x \right ]}{h}
= 5\lim_{h \rightarrow 0}\frac{[\sec(x + h) – \sec x]}{h} + 4 \lim_{h \rightarrow 0}\frac{[\cos(x + h) – \cos x]}{h}
= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (x + h) – \cos x \right ]
= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{\cos x – \cos (x + h)}{\cos x \cos (x + h)} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h}\left [ \cos x \cos h – \sin x \sin h – \cos x \right ]
\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{x + x + h}{2} \right )\sin \left ( \frac{x – x – h}{2} \right )}{cos(x + h)} \right ] + 4 \lim_{h\rightarrow 0}\frac{1}{h} [-\cos x (1 – \cos h) – \sin x sin h]
\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{ 2x + h}{2} \right )\sin \left ( \frac{ – h}{2} \right )}{cos(x + h)} \right ] + 4\left [ – \cos x \lim_{h \rightarrow 0} \frac{(1 – \cos h)}{h} – \sin x \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ]
= \frac{5}{\cos x}. \lim_{h \rightarrow 0} \left [ \frac{\sin\left ( \frac{2x + h}{2} \right ). \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}}{\cos (x + h)}\right ] + 4 [(-\cos x).(0) – (\sin x). 1]
= \frac{5}{\cos x}. \left [ \lim_{h \rightarrow 0} \frac{\sin\left ( \frac{2x + h}{2} \right )}{\cos (x + h)}. \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ] + 4 \sin x
= \frac{5}{\cos x}. \frac{\sin x}{\cos x}. 1 – 4 \sin x
= 5 \sec x \tan x – 4 \sin x
(iv)
Let f(x) = cosec x
According to first principle
f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}
f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin(x + h)}{ \sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ). \sin \left ( \frac{x – x – h}{2} \right ) }{\sin (x + h) . \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ). \sin \left ( \frac{- h}{2} \right ) }{\sin (x + h) . \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )} }{\sin (x + h) . \sin x}
= \lim_{h \rightarrow 0} \left ( \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). }{\sin (x + h) . \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}
= \left ( \frac{-\cos x}{\sin x \sin x} \right ).1
= – cosec x cot x
(v)
Let f(x) = 3cot x + 5cosec x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{3 \cot (x + h) + 5 \csc (x + h) – 3 \cot x – 5 \csc x}{h}
= 3 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] + 5 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \csc x] ….. (1)
Now,
\lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos (x + h)}{\sin (x + h)} – \frac{\cos x}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos(x + h)\sin x – \cos x \sin (x + h)}{\sin x sin(x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x – x – h)}{\sin x sin(x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (- h)}{\sin x sin(x + h)} \right ]
= - \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ). \left ( \lim_{h \rightarrow 0} \frac{1}{\sin x. \sin (x + h)} \right )
= - 1 . \frac{1}{\sin x . \sin (x + 0)} = \frac{-1}{\sin^{2}x} = -\csc ^{2}x ……. (2)
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin (x + h) }{\sin x \sin (x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ) . \sin \left ( \frac{x – x – h}{2} \right )}{\sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ) . \sin \left ( \frac{- h}{2} \right )}{\sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{ -\cos \left ( \frac{ 2x + h}{2} \right ) . \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}}{\sin (x + h) \sin x}
= \lim_{h \rightarrow 0} \left ( \frac{ -\cos \left ( \frac{ 2x + h}{2} \right )} {\sin (x + h) \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}
= \left ( \frac{- \cos x }{\sin x \sin x} \right ) . 1
= – \csc x \cot x ………. (3)
From eqn (1), (2), and (3), we obtain
f'(x) = – 3 \csc^{2} x – 5 \csc x \cot x
(vi)
Let f(x) = 5sin x – 6cos x + 7
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \sin(x + h) – 6 \cos (x + h) + 7 – 5 \sin x + 6 \cos x – 7 \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \left \{ \sin(x + h) – \sin x \right \} – 6 \left \{ \cos (x + h) – \cos x \right \} \right ]
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sin(x + h) – \sin x \right ] – 6 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \cos (x + h) – \cos x \right ]
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{x + x + h}{2} \right ) \sin \frac{x + h – x}{2} \right ] – 6 \lim_{h \rightarrow 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h}
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{ 2x + h}{2} \right ) \sin \frac{ h }{2} \right ] – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h ) – \sin x \sin h }{h} \right ]
= 5 \lim_{h \rightarrow 0} \left ( \cos\left ( \frac{ 2x + h}{2} \right ) \frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ) – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h )}{h} – \frac {\sin x \sin h }{h} \right ]
= 5 \left [ \lim_{h \rightarrow 0} \cos\left ( \frac{ 2x + h}{2} \right ) \right ] \left [ \lim_{\frac{h}{2} \rightarrow 0}\frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ] – 6 \left [ (- \cos x ) \left ( \lim_{h \rightarrow 0} \frac{ (1 – \cos h )}{h} \right ) – \sin x \lim_{h \rightarrow 0} \left ( \frac {\sin h }{h} \right ) \right ]
= 5 \cos x. 1 – 6 [(-\cos x). (0) – \sin x.1]
= 5cos x + 6 sin x
(vii)
Let f (x) = 2 tan x – 7 sec x
Accordingly to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \tan (x + h) – 7 \sec (x + h) – 2 \tan x + 7 \sec x\right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \left \{ \tan (x + h) – \tan x \right \} – 7 \left \{ \sec (x + h) – \sec x \right \} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \tan (x + h) – \tan x \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sec (x + h) – \sec x \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h)}{\cos (x + h)} – \frac{\sin x}{\cos x} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h) \cos x – \sin x \cos (x + h) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos (x + h) }{\cos x \cos (x + h)} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{ \sin (x + h – x) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \sin \left ( \frac{x – x – h}{2} \right ) \right )}{\cos x \cos (x + h)} \right ]
= 2 \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ) \left ( \lim_{h \rightarrow 0} \frac{ 1 }{ \cos x \cos (x + h)} \right ) – 7 \left( \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ) \left ( \lim_{h \rightarrow 0} \frac{ \sin \left ( \frac{ 2x + h}{2} \right ) }{\cos x \cos (x + h)} \right )
= 2 . 1 . \frac{1}{\cos x \cos x} – 7 . 1 \left ( \frac{\sin x}{\cos x \cos x} \right )
= 2 \sec^{2} x – 7 \sec x \tan x
BYJU'S Exercise 13.1
Q1.
Evaluate limx→3x+5 limit:
Soln:
limx→3x+5=3+5=8
Q2.
Evaluate limx→π(x–227)
Soln:
limx→π(x–227)=(π–227)
Q3.
Evaluate limr→lπr2
Soln:
limr→lπr2=π(l)2=π
Q4.
Evaluate limx+44x+6x–3
Soln:
limx+44x+6x–3=4(4)+64–3=16+61=221=22
Q5.
Evaluate limx→−1x12+x7+1x–1
Soln:
limx→−1x12+x7+1x–1=(−1)12+(−1)7+1−1–1=1–1+1−2=−12
Q6.
Evaluate limx→0(x+3)5–3x
Soln:
limx→0(x+3)5–3x
Put x + 3 = y so that y→1asx→0
Accordingly, limx→0(x+3)5–3x=limy→1y5–3y–3=limy→1y5–35y–3=5.35–1=405∴limx→0(x+3)5–3x=405
Q7.
Evaluate limx→23x2–x–10x2–4
Soln:
At x = 2, the given rational function has the value of 0/ 0
limx→23x2–x–10x2–4=limx→2(x–2)(3x+5)(x–2)(x+2)=limx→23x+5x+2=3(2)+52+2=114
Q8.
Evaluate limx→3x4–812x2–5x–3
Soln:
At x = 2, the given rational function has the value of 0/ 0
limx→3x4–812x2–5x–3=limx→3(x−3)(x+3)(x2+9)(x–3)(2x+1)=limx→3(x+3)(x2+9)2x+1=(3+3)(32+9)2(3)+1=6×187=1087
Q9.
Evaluate limx→0ax+bcx+1
Soln:
limx→0ax+bcx+1=a(0)+bc(0)+1=b
Q10.
Evaluate limz→1z13–1z16–1
Soln:
limz→1z13–1z16–1
At z = 1, the given function has the value of 0/ 0
Put z16=xsothatz→1asx→1Accordingly,limz→1z13–1z16–1=limx→1x2–1x–1=limx→1x2–12x–1=2.12–1=2∴limz→1z13–1z16–1=2
Q11.
Evaluate limx→1ax2+bx+ccx2+bx+a,a+b+c≠0
Soln:
limx→1ax2+bx+ccx2+bx+a=a(1)2+b(1)+cc(1)2+b(1)+a=a+b+ca+b+c=1
Q12.
limx→21x+12x+2
Soln:
limx→21x+12x+2
At x = -2, the given function has the value of 0/ 0
Now,limx→21x+12x+2=limx→−2(2+x2x)x+2=limx→−212x=12(−2)=−14
Q13.
Evaluate limx→0sinaxbx
Soln:
limx→0sinaxbx
At x = 0, the given function has the value of 0/ 0
Now,limx→0sinaxbx=limx→0sinaxax×axbx=limx→0(sinaxax)×(ab)=ablimax→0(sinaxax)=ab×1=ab
Q14.
Evaluate limx→0sinaxsinbx,a,b≠0
Soln:
limx→0sinaxsinbx,a,b≠0
At x = 0, the given function has the value of 0/ 0
Now,limx→0sinaxsinbx=limx→0(sinaxax)×ax(sinbxbx)×bx=(ab)×limax→0(sinaxax)limbx→0(sinbxbx)=(ab)×11=ab
Q15.
Evaluate limx→πsin(π–x)π(π–x)
Soln:
limx→πsin(π–x)π(π–x)
It seems x→π⇒(π–x)→0 ∴limx→πsin(π–x)π(π–x)=1πlim(z–x)–1sin(π–x)(π–x)=1π×1=1π
Q16.
Evaluate limx→0cosxπ–x
Soln;
limx→0cosxπ–x=cos0π–0=1π
Q17.
Evaluate limx→0cos2x–1cosx–1
Soln:
limx→0cos2x–1cosx–1
At x = 0, the given function has the value of 0/ 0
Now,limx→0cos2x–1cosx–1=limx→01–2sin2x–11–2sin2x2–1=limx→0sin2xsin2x2=limx→0(sin2xx2)×x2⎛⎝sin2x2(x2)2⎞⎠×x24=4limx→0(sin2xx2)limx→0⎛⎝sin2x2(x2)2⎞⎠ =4(limx→0sinxx)2(limx2→0sinx2x2)2=41212=4
Q18.
Evaluate limx→0ax+xcosxbsinx
Soln:
limx→0ax+xcosxbsinx
At x = 0, the given function has the value of 0/ 0
Nowlimx→0ax+xcosxbsinx=1blimx→0x(a+cosx)sinx=1blimx→0(xsinx)×limx→0(a+cosx)=1b×1(limx→0sinxx)×limx→0(a+cosx)=1b×(a+cos0)=a+1b
Q19.
Evaluate limx→0xsecx
Soln:
limx→0xsecx=limx→0xcosx=0cos0=01=0
Q20.
Evaluate limx→0sinax+bxax+sinbxa,ba+b≠0
Soln:
At x =0, the given function has the value of 0/ 0
Now,limx→0sinax+bxax+sinbxlimx→0(sinaxax)ax+bxax+bx(sinbxbx)= =(limax→0sinaxax)×limx→0(ax)+limx→0bxlimx→0ax+limx→0bx(limbx→0sinbxbx)=limx→0(ax)+limx→0bxlimx→0ax+limx→0bx=limx→0(ax+bx)limx→0(ax+bx)=limx→0(1)=1
Q21.
Evaluate limx→0(cscx–cotx)
Soln:
At x =0, the given function has the value of ∞ – ∞
Now,limx→0(cscx–cotx)=limx→0(1sinx–cosxsinx)=limx→0(1–cosxsinx)=limx→0(1–cosxsinx)(sinxx)=limx→01–cosxxlimx→0sinxx=01=0
Q22.
limx→πxtan2xx–π2
Soln:
limx→πxtan2xx–π2
At x=π2, the given function has the value of 0/ 0
Now,putx–π2=ysothatx→π2,y→0∴limx→π2tan2xx–π2=limy→0tan2(y+π2)y=limy→0tan(π+2y)y=limy→0tan2yy=limy→0sin2yycos2y=limy→0(sin2y2y×2cos2y)=(lim2y→0sin2y2y)×limy→0(2cos2y)=1×2cos0=1×21=2
Q23.
Find limx→0f(x)andlimx→0f(x),wheref(x)={2x+33(x+1)x≤0x>0
Soln:
Given function f(x)={2x+33(x+1)x≤0x>0
limx→0f(x)=limx→0[2x+3]=2(0)+3=3
limx→0f(x)=limx→03(x+1)=3(0+1)=3
∴limx→0f(x)=limx→0f(x)=limx→0f(x)=3
limx→1f(x)=limx→13(x+1)=3(1+1)=6
limx→1f(x)=limx→13(x+1)=3(1+1)=6
∴limx→1f(x)=limx→1f(x)=limx→1f(x) = 6
Q24.
Find limx→1f(x),wheref(x)={x2–1,−x2–1,x≤1x≥1
Soln:
Given function is
f(x)={x2–1,−x2–1,x≤1x≥1 limx→1−f(x)=limx→1[x2–1]=12–1=1–1=0 limx→1+f(x)=limx→1[−x2–1]=−12–1=1–1=0
We observed that limx→1−f(x)≠limx→1+f(x)
Hence, limx→1f(x) doesn’t exist.
25.
Evaluate limx→0f(x),wheref(x)={|x|x,0,x≠0x=0
Soln:
Given function f(x)={|x|x,0,x≠0x=0
When |x|=−x limx→0−f(x)=limx→0−[|x|x]
= limx→0(−xx)
= limx→0(−1)
= -1
When |x|=x limx→0+f(x)=limx→0+[|x|x]
= limx→0[−xx]
= limx→0(1)
= 1
We observe that limx→0−f(x)≠limx→0+f(x)
Hence, limx→0f(x) doesn’t exist.
Q26.
Find limx→0f(x),wheref(x)={x|x|,0,x≠0x=0
Soln:
f(x) = {x|x|,0,x≠0x=0
When |x|=−x limx→0−f(x)=limx→0−[x|x|] =limx→0[x–x] =limx→0(–1)
= -1
When |x|=x limx→0+f(x)=limx→0+[x|x|] =limx→0[xx] =limx→0(1)
= 1
We observe that limx→0−f(x)≠limx→0+f(x)
Hence, limx→0f(x) doesn’t exist.
Q27.
Find limx→5f(x),wheref(x)=|x|–5
Soln:
Given function is f(x)=|x|–5
When x>0,|x|=x limx→5−f(x)=limx→5−[|x|–5]
= limx→5(x–5)
= 5 – 5
= 0
When x>0,|x|=x limx→5+f(x)=limx→5+(|x|–5)
= limx→5(x–5)
= 5 – 5
= 0
∴limx→5−f(x)=limx→5+f(x)=0
Hence, limx→5f(x)=0
Q28.
Suppose f(x)=⎧⎩⎨⎪⎪a+bx,4,b–ax,ifx<1ifx=0,ifx>1
And limx→1f(x)=f(1) what can be the values of b and a?
Soln:
Given function
f(x)=⎧⎩⎨⎪⎪a+bx,4,b–ax,ifx<1ifx=0,ifx>1 limx→1−f(x)=limx→1(a+bx)=a+b limx→1−f(x)=limx→1(b–ax)=b–a
f(1) = 4
Given that limx→1f(x)=f(1).
∴limx→1−f(x)=limx→1+f(x)=limx→1f(x)=f(1)
=> a+ b = 4 and b – a = 4
Solving both the equations, we get a = 0 and b = 4.
Hence, b and a are 4 and 0
Q29.
A function f(x)=(x–a1)(x–a2)....(x–an) is define by fixed real numbers a1, a2, . . . . an.
Find limx→a1f(x). For some a≠a1,a2,....,ancomputelimx→anf(x)
Soln:
Given function = f(x)=(x–a1)(x–a2)...(x–an) limx→a1f(x)=limx→a1[(x–a1)(x–a2)...(x–an)]
=[limx→a1(x–a1)][limx→a1(x–a2)]...[limx→a1(x–an)]
=(a1–a1)(a1–a2)....(a1–an)=0
∴limx→a1]f(x)=0
Now, limx→a1]f(x)=limx→a[(x–a1)(x–a2)...(x–an)]
= =limx→a[x–a1][x–a2]...[x–an]
(a1–a1)(a1–a2)....(a1–an) ∴limx→af(x)=(a1–a1)(a1–a2)....(a1–an)
Q30.
If f(x)=⎧⎩⎨⎪⎪|x|+1,0,|x|–1,x≤0x=0x>0
limx→af(x) will exist for what values?
Soln:
Given function:
f(x)=⎧⎩⎨⎪⎪|x|+1,0,|x|–1,x≤0x=0x>0
When a = 0
limx→0−f(x)=limx→0−(|x|+1) =limx→0(–x+1) [if x < 0, |x| = -x]
= -0 + 1
= 1
limx→0+f(x)=limx→0+(|x|–1) =limx→0(x–1) [if x > 0, |x| = x]
= 0 – 1
= -1
We have, limx→0−f(x)≠limx→0+f(x) ∴limx→0f(x) doesn’t exist.
When a < 0
limx→a−f(x)=limx→a−(|x|+1) =limx→a(−x+1)[x<a<0⇒|x|=–x]
= -a + 1
limx→a+f(x)=limx→a+(|x|+1) =limx→a(−x+1)[x<a<0⇒|x|=–x]
= – a + 1
∴limx→a−=limx→a+=–a+1
Hence, at x = a limit of f(x) exist, where a > 0
Thus, limx→af(x)existforalla≠0
Q31.
Evaluate limx→1f(x)
If f(x) satisfies, limx→1f(x)–2x2–1=π
Soln:
limx→1f(x)–2x2–1=π ⇒limx→1(f(x)–2)limx→1(x2–1)=π ⇒limx→1(f(x)–2)=πlimx→1(x2–1) ⇒limx→1(f(x)–2)=π(12–1) ⇒limx→1(f(x)–2)=0 ⇒limx→1f(x)–limx→12=0 ⇒limx→1f(x)–2=0 ∴limx→1f(x)=2
Q32.
If f(x)=⎧⎩⎨⎪⎪ax2+b,bx+a,bx3+ax<00≤x≤1x>1
For what values of a and b does limx→0f(x)andlimx→1f(x) exist?
Soln:
Given function
f(x)=⎧⎩⎨⎪⎪ax2+b,bx+a,bx3+ax<00≤x≤1x>1 limx→0−f(x)=limx→0(ax2+b)
= a(0)2 + b
= b
limx→0+f(x)=limx→0(bx+a)
= b(1) + a
= a + b
∴limx→1−f(x)=limx→1+f(x)=limx→1f(x)
Thus, for integral values of n and m limx→1f(x) exist.
Exercise-13.2
Q1. Find derivative for x2 – 2 at x = 10
Soln:
Let f(x) = x2 – 2
Accordingly,
f′(10)=limh→0f(10+h)–f(10)h
=limh→0[(10+h)2–2]–(102–2)h
=limh→0102+2.10.h+h2–2–102+2h
=limh→020h+h2h
=limh→0(20+h)=(20+0)=0
Hence, derivative for x2 – 2 at x = 10 is 20
Q2. Find derivative
99x at x = 100
Soln:
Let f(x) = 99x,
Accordingly,
f′(100)=limh→0f(100+h)–f(100)h
=limh→099(100+h)–99(100)h
=limh→099×100+99h–99×100h
=limh→099hh
=limh→0(99h)=99
Hence, derivative for 99x at x = 100 is 99
Q3. Find derivative
x at x = 1
Soln:
Let f(x) = x
Accordingly,
f′(1)=limh→0f(1+h)–f(1)h
=limh→0(1+h)–1h
=limh→0hh
=limh→0(1)
= 1
Hence, derivative for x at x = 1 is 1
Q4. Using first principle find derivative
(i) x3 – 27
(ii) (x – 1) (x – 2)
(iii) 1x2
(iv) x+1x–1
Soln:
(i) Let f(x) = x3 – 27
From first principle,
f′(x)=limh→0f(x+h)–f(x)h
=limh→0[(x+h)3–27]–(x3–27)h
=limh→0x3+h3+3x2h+3xh2–x3h
=limh→0h3+3x2h+3xh2h
=limh→0(h2+3x2+3xh)
= 0 + 3x2 + 0 = 3x2
(ii) Let f(x) = (x – 1)(x – 2)
From first principle,
f′(x)=limh→0f(x+h)–f(x)h
=limh→0(x+h–1)(x+h–2)–(x–1)(x–2)h
=limh→0(x2+hx–2x+hx+h2–2h–x–h+2)–(x2–2x–x+2)h
=limh→0(hx+hx+h2–2h−h)h
=limh→0(2hx+h2–3h)h
=limh→0(2x+h–3)
(2x+h–3)
= 2x – 3
(iii) Let f(x)=1x2
From first principle
f′(x)=limh→0f(x+h)–f(x)h
=limh→01(x+h)2–1x2h
=limh→01h[x2–(x+h)2x2(x+h)2]
=limh→01h[x2–x2–h2–2hxx2(x+h)2]
=limh→01h[–h2–2hxx2(x+h)2]
=limh→0[–h–2hxx2(x+h)2]
0–2xx2(x+0)2=−2x3
(iv) Let f(x)=x+1x–1
From first principle,
=limh→0(x+h+1x+h–1–x+1x–1)h
=limh→01h[(x–1)(x+h+1)–(x+1)(x+h–1)(x–1)(x+h–1)]
=limh→01h[(x2+hx+x–x–h–1)–(x2+hx–x+x+h–1)(x−1)(x+h+1)]
=limh→01h[−2h(x–1)(x+h–1)]
=limh→0[−2(x–1)(x+h–1)]
=−2(x–1)(x–1)=−2(x–1)2
Q5. Prove f′(1)=100f′(0)
For the function f(x)=x100100+x9999+…+x22+x+1
Soln:
Given function
f(x)=x100100+x9999+…+x22+x+1
ddxf(x)=ddx[x100100+x9999+…+x22+x+1]
ddxf(x)=ddx(x100100)+ddx(x9999)+…+ddx(x22)+ddx(x)+ddx(1)
Using theorem ddx(xn)=nxn–1, we get
ddxf(x)=100x99100+99x9899+....+2x2+1+0
=x99+x98+….+x+1
∴ ddxf(x)=x99+x98+….+x+1
At x = 0
f'(0) = 1
At x = 1,
f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100
Hence, f’(1) = 100 x f1 (0)
Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an
Soln:
Let f(x)=xn+axn–1+a2xn–2+...+an–1x+an
f′(x)=ddx(xn+axn–1+a2xn–2+...+an–1x+an)
=ddx(xn)+addx(xn–1)+a2ddx(xn–2)+...+an–1ddx(x)+anddx(1)
Using ddxxn=nxn−1, we have
f′(x)=nxn–1+a(n–1)xn–2+a2(n–2)xn−3+…+an–1+an(0)
=nxn–1+a(n–1)xn–2+a2(n–2)xn−3+…+an–1
Q7. Find the derivative for
(i) (x – m)(x – n)
(ii) (ax2 + b)2
(iii) x–ax–b
Soln:
(i) Let f(x) = (x – m)(x – n)
⇒f(x)=x2–(m+n)x+mn
f′(x)=ddx(x2–(m+n)x+mn)
=ddx(x2)–(m+n)ddx(x)+ddx(mn)
Using =ddx(xn)=nxn–1, we get
f′(x)=2x−(m+n)+0=2x–m–n
(ii) Let f(x) = (ax2 + b)2
⇒f(x)=a2x2+2abx+b2
f′(x)=ddx(a2x4+2abx2+b2)=a2ddx(x4)+2abddx(x2)+ddx(b2)
Using ddxxn=nxn–1, we have
f′(x)=a2(4x3)+2ab(2x)+b2(0)
= 4a2x3 + 4abx
= 4ax(ax2 + b)
(iii) Let f(x)=(x–a)(x–b)
f′(x)=ddx(x–ax–b)
Quotient rule,
f′(x)=(x–b)ddx(x–a)–(x–a)ddx(x−b)(x–b)2
=(x–b)(1)–(x–a)(1)(x–b)2
=x–b–x+a(x–b)2
=a–b(x–b)2
Q8. If a is constant find derivative of xn–anx–a
Soln:
Let xn–anx–a
⇒f′(x)=ddx(xn–anx–a)
By Question rule
⇒f′(x)=(x–a)ddx(xn–an)–(xn–an)ddx(x–a)(x–a)2
=(x−a)(nxn–1–0)–(xn–an)(x–a)2
=nxn–anxn–1–xn+an(x–a)2
Q9.
(i) 2x – \frac {3}{4}
(ii) (5x3 + 3x – 1) (x – 1)
(iii) x-3 (5 + 3x)
(iv) x5 (3 – 6x-9)
(v) x-4 (3 – 4x-5)
(vi) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1}
Soln.
(i) Let f(x) = 2x – \frac {3}{4}
f'(x) = \frac{d}{dx}\left ( 2x – \frac{3}{4} \right )
= 2\frac{d}{dx}\left ( x \right ) – \frac {d}{dx} \left ( \frac{3}{4} \right )
= 2 – 0
= 2
(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)
By Leibnitz product rule,
f'(x) = (5x^{3} + 3x – 1) \frac{d}{dx}(x – 1) + (x – 1) \frac{d}{dx}(5x^{3} + 3x – 1 )
= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)
= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)
= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3
= 20x3 – 15x2 + 6x – 4
(iii) Let f(x) = x-3 (5 + 3x)
f'(x) = x^{-3}\frac{d}{dx}(5 + 3x) + (5 + 3x)\frac{d}{dx}(x^{-3})
= x^{-3}(0 + 3) + (5 + 3x) (-3x^{-3 – 1} )
= x^{-3}( 3) + (5 + 3x) (-3x^{-4} )
= 3x-3 – 15x-4 – 9x-3
= -6x-3 – 15x-4
= -3 x ^{-3} \left ( 2 + \frac{5}{x} \right )
= \frac{-3 x ^{-3} }{x}\left ( 2x + 5 \right )
= \frac{-3 }{ x ^{4} }\left ( 2x + 5 \right )
(iv) let f(x) = x5 (3 – 6x-9)
From Leibnitz product rule,
f'(x) = x^{5} \frac{d}{dx}(3 – 6x^{-9}) + (3 – 6x^{-9}) \frac{d}{dx}(x^{5})
= x^{5} \left \{ 0 – 6 (-9)x^{-9 – 1} \right \} + (3 – 6x^{-9})(5x^{4})
= x^{5} (54x ^{-10}) + 15x^{4} – 30x^{-5}
= 54x-5 + 15x4 – 30x-5
= 24x-5 + 15x4
= = 15x^{4} + \frac{24}{x^{5}}
(v) Let f(x) = x-4 (3 – 4x-5)
From Leibnitz product rule,
f'(x) = x^{-4}\frac{d}{dx}\left ( 3 – 4x^{-5} \right ) + (3 – 4 x^{-5})\frac{d}{dx}(x^{-4})
= x^{-4}\left \{ 0 – 4 (-5)x^{-5 – 1} + (3 – 4x^{-5})(-4)x^{-4 -1} \right \}
= x-4 (20x-6) + (3 – 4x-5)(-4x-5)
= 36x-10 – 12x-5
= -\frac{12}{x^{5}} + \frac{36}{x^{10}}
(iv) Let f(x) = \frac{2}{x + 1} – \frac{x^{2}}{3x – 1} f'(x) = \frac{d}{dx}\left ( \frac{2}{x + 1} \right ) – \frac{d}{dx} \left ( \frac{x^{2}}{3x – 1} \right )
From quotient rule,
f'(x) = \left [ \frac{\left ( x + 1 \right ) \frac{d}{dx}(2) – 2 \frac{d}{dx} (x + 1) }{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1) \frac{d}{dx} (x^{2}) – x^{2} \frac{d}{dx}\left ( 3x – 1 \right )}{(3x – 1)^{2}} \right ]
= \left [ \frac{(x + 1)(0) – 2(1)}{(x + 1)^{2}} \right ] – \left [ \frac{(3x – 1)(2x) – (x^{2})(3)}{(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \left [ \frac{6x^{2} – 2x – 3x^{2}}{(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \left [ \frac{3x^{2} – 2x} {(3x – 1)^{2}} \right ]
= \frac{-2}{(x + 1)^{2}} – \frac{x(3x – 2)} {(3x – 1)^{2}}
Q10. Using first principle find derivative of \cos x
Soln:
Let f(x) = \cos x.
According to first principle
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{\cos (x + h) – \cos x}{h}
= \lim_{h \rightarrow 0} \left [\frac{\cos x \cos h – \sin x \sin h – \cos x}{h} \right ]
= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h) – \sin x \sin h }{h} \right ]
= \lim_{h \rightarrow 0} \left [\frac{ – \cos x (1 – \cos h)}{h} – \frac{\sin x \sin h }{h} \right ]
=- \cos x \left ( \lim_{h \rightarrow 0} \frac{1 – \cosh}{h} \right ) – \sin x\lim_{h \rightarrow 0}\left ( \frac{sin h}{h} \right )
= = – \cos x (0) – \sin x (1) \left [ \lim_{h \rightarrow 0} \frac{1 – cos h}{h} = 0 \; and \; \lim_{h \rightarrow 0}\frac{\sin h }{h} = 1 \right ]
= – sin x
f’(x) = – sin x
Q11.
(i) sin x cos x
(ii) sec x
(iii) 5 sec x + 4 cos x
(iv) cosec x
(v) 3cot x + 5cosec x
(vi) 5sin x – 6cos x + 7
(vii) 2tan x – 7sec x
Soln:
(i) Let f’(x) = sin x cos x
According to first principle,
f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) -f(x)}{h}
= \lim_{h \rightarrow 0}\frac{\sin (x + h) \cos(x + h) – \sin x \cos x}{h}
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2\sin (x + h) \cos (x + h) – 2 \sin x cos x \right ]
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ \sin 2(x + h) – \sin 2x \right ]
= \lim_{h \rightarrow 0}\frac{1}{2h} \left [ 2 \cos \frac{2x + 2h + 2x}{2} . \sin \frac{2x + 2h – 2x}{2} \right ]
= \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos \frac{ 4x + 2h }{2} \sin \frac{ 2h }{2} \right ]
\lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (2x + h ) \sin h \right ]
\lim_{h \rightarrow 0} \cos (2x + h ) \lim_{h \rightarrow 0} \frac{\sin h}{h}
= cos(2x + 0). 1
= cos 2x
(ii) Let f(x) = sec x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{\sec (x + h) – \sec x}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos(x + h)}{\cos x \cos(x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \right ) \sin \left ( \frac{x – x – h}{2} \right )}{\cos (x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{2x + h}{2} \right ) \sin \left ( \frac{- h}{2} \right )}{\cos (x + h)} \right ]
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\left [ \sin \left ( \frac{2x + h}{2} \right ) \frac{\sin \left ( \frac{h}{2}\right )}{\left (\frac{h}{2}\right )} \right ]}{\cos (x + h)}
= \frac{1}{\cos x} \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\left ( \frac{h}{2} \right )}. \lim_{h \rightarrow 0}\frac{\sin\left ( \frac{2x + h}{2} \right ) }{\cos (x + h)}
= \frac{1}{\cos x} . 1 . \frac{\sin x}{ \cos x}
= \sec x \tan x
(iii)
Let f(x) = 5 sec x + 4 cos x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
\lim_{h \rightarrow 0}\frac{5\sec (x + h) + 4 \cos(x + h) – \left [ 5 \ sec x + 4 \ cos x \right ]}{h}
= 5\lim_{h \rightarrow 0}\frac{[\sec(x + h) – \sec x]}{h} + 4 \lim_{h \rightarrow 0}\frac{[\cos(x + h) – \cos x]}{h}
= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h} \left [ \cos (x + h) – \cos x \right ]
= 5\lim_{h \rightarrow 0}\frac{1}{h}\left [ \frac{\cos x – \cos (x + h)}{\cos x \cos (x + h)} \right ] + 4 \lim_{h \rightarrow 0}\frac{1}{h}\left [ \cos x \cos h – \sin x \sin h – \cos x \right ]
\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{x + x + h}{2} \right )\sin \left ( \frac{x – x – h}{2} \right )}{cos(x + h)} \right ] + 4 \lim_{h\rightarrow 0}\frac{1}{h} [-\cos x (1 – \cos h) – \sin x sin h]
\\= \frac{5}{\cos x} \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2\sin \left ( \frac{ 2x + h}{2} \right )\sin \left ( \frac{ – h}{2} \right )}{cos(x + h)} \right ] + 4\left [ – \cos x \lim_{h \rightarrow 0} \frac{(1 – \cos h)}{h} – \sin x \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ]
= \frac{5}{\cos x}. \lim_{h \rightarrow 0} \left [ \frac{\sin\left ( \frac{2x + h}{2} \right ). \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}}{\cos (x + h)}\right ] + 4 [(-\cos x).(0) – (\sin x). 1]
= \frac{5}{\cos x}. \left [ \lim_{h \rightarrow 0} \frac{\sin\left ( \frac{2x + h}{2} \right )}{\cos (x + h)}. \lim_{h \rightarrow 0} \frac{\sin \left ( \frac{h}{2} \right )}{\frac{h}{2}}\right ] + 4 \sin x
= \frac{5}{\cos x}. \frac{\sin x}{\cos x}. 1 – 4 \sin x
= 5 \sec x \tan x – 4 \sin x
(iv)
Let f(x) = cosec x
According to first principle
f'(x) = \lim_{h \rightarrow 0}\frac{f(x + h) – f(x)}{h}
f'(x) = \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin(x + h)}{ \sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ). \sin \left ( \frac{x – x – h}{2} \right ) }{\sin (x + h) . \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ). \sin \left ( \frac{- h}{2} \right ) }{\sin (x + h) . \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )} }{\sin (x + h) . \sin x}
= \lim_{h \rightarrow 0} \left ( \frac{ – \cos \left ( \frac{ 2x + h}{2} \right ). }{\sin (x + h) . \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}
= \left ( \frac{-\cos x}{\sin x \sin x} \right ).1
= – cosec x cot x
(v)
Let f(x) = 3cot x + 5cosec x
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{3 \cot (x + h) + 5 \csc (x + h) – 3 \cot x – 5 \csc x}{h}
= 3 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x] + 5 \lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \csc x] ….. (1)
Now,
\lim_{h \rightarrow 0} \frac{1}{h} [\cot (x + h) – \cot x]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos (x + h)}{\sin (x + h)} – \frac{\cos x}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos(x + h)\sin x – \cos x \sin (x + h)}{\sin x sin(x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x – x – h)}{\sin x sin(x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (- h)}{\sin x sin(x + h)} \right ]
= - \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ). \left ( \lim_{h \rightarrow 0} \frac{1}{\sin x. \sin (x + h)} \right )
= - 1 . \frac{1}{\sin x . \sin (x + 0)} = \frac{-1}{\sin^{2}x} = -\csc ^{2}x ……. (2)
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \csc (x + h) – \csc x \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\sin (x + h)} – \frac{1}{\sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin x – \sin (x + h) }{\sin x \sin (x + h)} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{x + x + h}{2} \right ) . \sin \left ( \frac{x – x – h}{2} \right )}{\sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{2 \cos \left ( \frac{ 2x + h}{2} \right ) . \sin \left ( \frac{- h}{2} \right )}{\sin (x + h) \sin x} \right ]
= \lim_{h \rightarrow 0} \frac{ -\cos \left ( \frac{ 2x + h}{2} \right ) . \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}}{\sin (x + h) \sin x}
= \lim_{h \rightarrow 0} \left ( \frac{ -\cos \left ( \frac{ 2x + h}{2} \right )} {\sin (x + h) \sin x} \right ) \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \left ( \frac{ h}{2} \right )}{\left ( \frac{ h}{2} \right )}
= \left ( \frac{- \cos x }{\sin x \sin x} \right ) . 1
= – \csc x \cot x ………. (3)
From eqn (1), (2), and (3), we obtain
f'(x) = – 3 \csc^{2} x – 5 \csc x \cot x
(vi)
Let f(x) = 5sin x – 6cos x + 7
According to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \sin(x + h) – 6 \cos (x + h) + 7 – 5 \sin x + 6 \cos x – 7 \right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 5 \left \{ \sin(x + h) – \sin x \right \} – 6 \left \{ \cos (x + h) – \cos x \right \} \right ]
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sin(x + h) – \sin x \right ] – 6 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \cos (x + h) – \cos x \right ]
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{x + x + h}{2} \right ) \sin \frac{x + h – x}{2} \right ] – 6 \lim_{h \rightarrow 0} \frac{\cos x \cos h – \sin x \sin h – \cos x}{h}
= 5 \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \cos\left ( \frac{ 2x + h}{2} \right ) \sin \frac{ h }{2} \right ] – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h ) – \sin x \sin h }{h} \right ]
= 5 \lim_{h \rightarrow 0} \left ( \cos\left ( \frac{ 2x + h}{2} \right ) \frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ) – 6 \lim_{h \rightarrow 0} \left [ \frac{ – \cos x (1 – \cos h )}{h} – \frac {\sin x \sin h }{h} \right ]
= 5 \left [ \lim_{h \rightarrow 0} \cos\left ( \frac{ 2x + h}{2} \right ) \right ] \left [ \lim_{\frac{h}{2} \rightarrow 0}\frac{ \sin \frac{ h }{2}}{\frac{ h }{2}} \right ] – 6 \left [ (- \cos x ) \left ( \lim_{h \rightarrow 0} \frac{ (1 – \cos h )}{h} \right ) – \sin x \lim_{h \rightarrow 0} \left ( \frac {\sin h }{h} \right ) \right ]
= 5 \cos x. 1 – 6 [(-\cos x). (0) – \sin x.1]
= 5cos x + 6 sin x
(vii)
Let f (x) = 2 tan x – 7 sec x
Accordingly to first principle,
f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) – f(x)}{h}
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \tan (x + h) – 7 \sec (x + h) – 2 \tan x + 7 \sec x\right ]
= \lim_{h \rightarrow 0} \frac{1}{h} \left [ 2 \left \{ \tan (x + h) – \tan x \right \} – 7 \left \{ \sec (x + h) – \sec x \right \} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \tan (x + h) – \tan x \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \sec (x + h) – \sec x \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h)}{\cos (x + h)} – \frac{\sin x}{\cos x} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{1}{\cos (x + h)} – \frac{1}{\cos x} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\sin (x + h) \cos x – \sin x \cos (x + h) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{\cos x – \cos (x + h) }{\cos x \cos (x + h)} \right ]
= 2 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{ \sin (x + h – x) }{ \cos x \cos (x + h)} \right ] – 7 \lim_{h \rightarrow 0} \frac{1}{h} \left [ \frac{-2 \sin \left ( \frac{x + x + h}{2} \sin \left ( \frac{x – x – h}{2} \right ) \right )}{\cos x \cos (x + h)} \right ]
= 2 \left ( \lim_{h \rightarrow 0} \frac{\sin h}{h} \right ) \left ( \lim_{h \rightarrow 0} \frac{ 1 }{ \cos x \cos (x + h)} \right ) – 7 \left( \lim_{\frac{h}{2} \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \right ) \left ( \lim_{h \rightarrow 0} \frac{ \sin \left ( \frac{ 2x + h}{2} \right ) }{\cos x \cos (x + h)} \right )
= 2 . 1 . \frac{1}{\cos x \cos x} – 7 . 1 \left ( \frac{\sin x}{\cos x \cos x} \right )
= 2 \sec^{2} x – 7 \sec x \tan x
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