# Class 11 RD Sharma Solutions – Chapter 30 Derivatives – Exercise 30.3

• Last Updated : 16 May, 2021

### Question 1. Differentiate f(x) = x4 – 2sinx + 3cosx with respect to x.

Solution:

Given that, f(x) = x4 – 2sinx + 3cosx

Now, differentiate w.r.t. x, we get

⇒ d(x4 – 2sinx + 3cosx) / dx

⇒ d(x4)/dx – 2.d(sinx)/dx + 3.d(cosx)/dx

⇒ 4x3 – 2cosx – 3sinx.

### Question 2. Differentiate f(x) = 3x + x3 + 33 with respect to x.

Solution:

Given that, f(x) = 3x + x3 + 33

Now, differentiate w.r.t. x, we get

d(3x + x3 + 33) / dx

⇒ d(3x)/dx + d(x3)/dx + d(33)/dx

⇒ 3x log3 + 3x2 + 0    [As we know, d(ax)/dx = ax loga]

⇒ 3x log3 + 3x2

### Question 3. Differentiate f(x) = x3/3 – 2√x + 5/x2 with respect to x.

Solution:

Given that, f(x) = x3/3 – 2√x + 5/x2

Now, differentiate w.r.t. x, we get

d(x3/3 – 2√x + 5/x2) / dx

⇒ 1.d(x3)/3dx – 2d(√x)/dx + 5d(x-2)/dx

⇒ 1/3.3x2 – 2.1/2.1/√x + 5(-2) x-3

⇒ x2 – x-1/2 – 10x-3

⇒ x2 – 1/√x – 10/x3

### Question 4. Differentiate f(x) = exloga + ealogx + ealoga with respect to x.

Solution:

Given that, f(x) = exloga + ealogx + ealoga

Now, differentiate w.r.t. x, we get

⇒ d(exloga + ealogx + ealoga)

⇒ d(exloga)/dx + d(ealogx)/dx + d(ealoga)/dx

⇒ exloga.loga + ealogx.a/x + 0       [As we know, ealoga is constant]

⇒ loga.exloga + a/x.ealogx

⇒ loga.ax + a/x xa                [Here, ax can be written as a exloga]

⇒ ax loga + axa-1

### Question 5. Differentiate f(x) = (2x2 + 1)(3x + 2) with respect to x.

Solution:

Given that, f(x) = (2x2 + 1)(3x + 2)

Now, differentiate w.r.t. x, we get

d(2x2 + 1)(3x + 2)/dx

⇒ (3x + 2)d(2x2 + 1)/dx + (2x2 + 1)d(3x + 2)/dx

⇒ (3x + 2)(4x+0) + (2x2 + 1)(3+ 0)

⇒ (12x2 + 8x + 6x2 + 3)

⇒ 18x2 + 8x + 3.

### Question 6. Differentiate f(x) = log3x + 3logex + 2tanx with respect x.

Solution:

Given that, f(x) = log3x + 3logex + 2tanx

Now, differentiate w.r.t. x, we get

d( log3x + 3logex + 2tanx)/dx

⇒ 1/log3 d(logx)/dx + 3.d(logex)/dx + 2.d(tanx)/dx

⇒ 1/log3 × 1/x + 3/x + 2sec2x

⇒ 1/xlog3 + 3/x + 2sec2x

### Question 7. Differentiate f(x) = (x + 1/x) (√x + 1/√x) with respect to x.

Solution:

Given that,  f(x) = (x + 1/x) (√x + 1/√x)

Now, differentiate w.r.t. x, we get

⇒ d((x + 1/x) (√x + 1/√x))/dx

⇒ (x + 1/x) d(√x + 1/√x)/dx  + (√x + 1/√x) d(x + 1/x)/dx

⇒ (x + 1/x) (1/2√x – 1/2x3/2) +  (√x + 1/√x) (1 – 1/x2)

⇒ {x/(2√x) – x/(2x3/2) + 1/2(x3/2)- 1/(2x5/2)} + {√x – √x/x2 + 1/√x – 1/x5/2}

⇒ (1.√x/2 – 1/2√x + 1/2x3/2 – 1/2x5/2 + √x – 1/x3/2 + 1/√x – 1/x5/2)

⇒ (3√x/2 + √x/2 – 1/2x3/2 – 3/2x5/2)

⇒ 3x1/2/2 + x-1/2/2 – x-3/2/2 – 3x-5/2/2

### Question 8. Differentiate f(x) = (√x + 1/√x)3 with respect to x.

Solution:

Given that, f(x) = (√x + 1/√x)3

Now, differentiate w.r.t. x, we get

⇒ d(√x + 1/√x)3 /dx

⇒ d(x3/2 + 3x.1/x + 3√x.1/x + 1/x3/2)/dx      [As we know that, (a + b)3 = a2+ 3a2b + 3ab2 + b3]

⇒ d(x3/2 + 3x1/2 + 3x-1/2 + x-3/2)/dx

⇒ 3x1/2/2 + 3x-1/2/2 + 3.(-1/2).x-3/2 – 3x-5/2/2

⇒ 3x1/2/2 – 3x-5/2/2 + 3x-1/2/2 – 3x-3/2/2.

### Question 9. Differentiate f(x) = 2x2 + 3x + 4 /x with respect to x.

Solution:

Given that, f(x) = 2x2 + 3x + 4 /x

Now, differentiate w.r.t. x, we get

⇒ d(2x2 + 3x + 4 /x) / dx

⇒d(2x2/x + 3x/x + 4/x) /dx

⇒ d(2x + 3 + 4x-1) / dx

⇒ 2- 4/x2

### Question 10. Differentiate f(x) = (x3+ 1) (x – 2) / x2 with respect x.

Solution:

Given that, f(x) = (x3+ 1) (x – 2) / x2

Now, differentiate w.r.t. x, we get

⇒ d{(x3 + 1) (x – 2) / x2} / dx

⇒ d{(x4 – 2x3 +x – 2)/ x2} / dx

⇒ d(x2 – 2x + x-1 – 2x-2) / dx

⇒ d(x2)/dx – 2d(x)/dx + d(x-1)/dx – 2d(x-2)/dx

⇒ 2x – 2 – 1/x2 + 4/x3

⇒ 2x – 2 – 1/x2 + 4/x3

### Question 11. Differentiate f(x) = acosx + bsinx + c / sinx with respect to x.

Solution:

Given that, f(x) = acosx + bsinx + c / sinx

Now, differentiate w.r.t. x, we get

⇒ d(acosx + bsinx + c / sinx) /dx

⇒ a.d(cosx)/dx(sinx) + b.d(1)/dx + c.d/dx(sinx)

⇒ a(-cosec2x) + 0 + c(-cosecx.cotx)

⇒ -acosec2x – c.cosecx.cotx

### Question 12. Differentiate f(x) = (2secx + 3cotx – 4tanx) with respect to x.

Solution:

Given that, f(x) = (2secx + 3cotx – 4tanx)

Now, differentiate w.r.t. x, we get

⇒ d(2secx + 3cotx – 4tanx) / dx

⇒ 2.d(2secx)/dx + 3.d(cotx)/dx – 4.d(tanx)/dx

⇒ 2secxtanx – 3cosec2x – 4sec2x

### Question 13. Differentiate f(x) = (a0xn + a1xn-1 + a2xn-2  + ……… + an-1x + an) with respect to x.

Solution:

Given that, f(x) = (a0xn + a1xn-1 + a2xn-2  + ……… + an-1x + an)

Now, differentiate w.r.t. x, we get

⇒ d(a0xn + a1xn-1 + a2xn-2 + ……… + an-1x + an) / dx

⇒ a0d(x)n/dx + a1d(x)n-1/dx + a2d(x)n-2/dx + ………. + an-1d(x)/dx + and(1)/dx

⇒ na0xn-1 + (n-1)a1xn-2 + ………. + an-1 + 0

⇒ na0xn-1+ (n-1)a1xn-2 + ……….. + an-1

### Question 14. Differentiate f(x) = 1/sinx + 2x+3 + 4/logx3 with respect to x.

Solution:

Given that, f(x) = 1/sinx + 2x+3 + 4/logx

Now, differentiate w.r.t. x, we get

⇒ d/dx (1/sinx + 2x+3 + 4/logx3)

⇒ d(cosecx)/dx + 23d(2x)/dx + 4/log3 × d(logx)/dx        [As we knwo that, logba = loga/logb]

⇒ -cosecx.cotx + 8 × 2log2 + 4/log3 × 1/x           [Since, d(ax)/dx = axloga]

⇒ -cosecx.cotx + 2x+3log2 + 4/xlog3

### Question 15. Differentiate f(x) = (x + 5)(2x – 1) / x with respect to x.

Solution:

Given that, f(x) = (x + 5)(2x – 1) / x

Now, differentiate w.r.t. x, we get

⇒ d/dx {(x + 5)(2x2 – 1)/x}

⇒ d/dx (2x3 + 10x2 – x – 5 / x)

⇒ d(2x2 + 10x – 1 – 5x-1)/dx

⇒ 2d(x2)/dx + 10d(x)/dx – d(1)/dx – 5d(x-1)/dx

⇒ 2.2x + 10 – 0 + 5/x2

⇒ 4x + 10 + 5/x

### Question 16. Differentiate f(x) = log(1/√x) + 5xa – 3ax + 3√x2 + 6(4√x-3) with respect to x.

Solution:

Given that, f(x) = log(1/√x) + 5xa – 3ax + 3√x2 + 6(4√x-3

Now, differentiate w.r.t. x, we get

⇒ d/dx {log(1/√x) + 5xa – 3ax + 3√x2 + 6(4√x-3)}

⇒ d(log(1/√x)/dx + 5d(xa)/dx – 3(ax) + d(3√x2)/dx + 6d(4√x-3)/dx

⇒ -1/2.1/x + 5axa-1 – 3axloga + 2x-1/3/3 + 6x-7/4(-3/4)

⇒ -1/2x + 5axa-1 – 3axloga + 2x-1/3/3 – 9x-7/4/2

### Question 17. Differentiate f(x) = cos(x + a) with respect to x.

Solution:

Given that, f(x) = cos(x + a)

Now, differentiate w.r.t. x, we get

⇒ d{cos(x + a)}/dx

⇒ d(cosx.cosa – sinx.sina)/dx

⇒ cosa.d(cosx)/dx – sina.d(sinx)/dx

⇒ cosa(-sinx) – sina(cosx)

⇒ cosx.sina + sinx.cosa

⇒ -(sinx.cosa + cosx.sina)

⇒ -sin(x + a)

### Question 18. Differentiate f(x) = cos(x – 2)/sinx with respect to x.

Solution:

Given that, f(x) = cos(x – 2)/sinx

Now, differentiate w.r.t. x, we get

⇒ d{cos(x – 2)/sinx)/dx

⇒ d{(cosx.cos2 + sinx.sin2)/sinx} / dx

⇒ cos2.d(cotx)/dx + sin2.d(1)/dx

⇒ -cos2.cosec2x + 0

⇒ -cosec2x.cos2

### Question 19. If y = {sin(x/2) + cos(x/2)}, find dy/dx at x = π/6.

Solution:

Given that, y = {sin(x/2) + cos(x/2)}  …..(1)

Find that dy/dx at x = π/6

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d{sin(x/2) + cos(x/2)}/dx

⇒ d{sin2(x/2) + cos2(x/2) + 2sin(x/2).cos(x/2)}/dx

⇒ d(1 + sinx)/dx       [As we know that sin2x + cos2x = 1]

⇒ 0 + cosx             [As we know that sin2x= 2sinx.cosx]

⇒ cosx

Now put x = π/6

⇒ cos(π/6)

⇒ √3/2

### Question 20. If y = (2 – 3cosx / sinx), find dy/dx at x = π/4.

Solution:

Given that, y = (2 – 3cosx / sinx)  ….(1)

Find that dy/dx at x = π/4

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d(2 – 3cosx / sinx) / dx

⇒ d(2cosecx – 3cotx) / dx

⇒ 2d(cosecx)/dx – 3d(cotx)/dx

⇒ -2cosecx.cotx + 3cosec2x

Now put x = π/4

⇒ -2cosec(π/4).cot(π/4) + 3cosec2(π/4)

⇒ -2√2 – 1 + 3.2

⇒ -2√2 + 6

⇒ 6 – 2√2

### Question 21. Find the slope of the tangent to the curve f(x) = 2x6 + x4 – 1 at x = 1.

Solution:

Given that f(x) = 2x6 + x4 – 1 at x = 1.

Find the slope of the tangent at a point x = 1

Now, differentiate w.r.t. x, we get

⇒ d(2x6 + x4 -1)/dx

⇒ 2dx6/dx + dx4/dx – d.1/dx

⇒ 12x5+ 4x3 – 0

⇒ 12x5 + 4x3

Now put x = 1

⇒ 12(1)5 + 4(1)3

⇒ 12 + 4

⇒ 16

Hence, the slope of the tangent to the curve f(x) at x = 1 is 16.

### Question 22. If y = √x/a + √a/x, prove that 2xy.dy/dx = (x/a – a/x)

Solution:

Given that, y = √x/a + √a/x

Prove that 2xy.dy/dx = (x/a – a/x)

Proof:

dy/dx = d(√x/a + √a/x)/dx

⇒ 1/√a.d(√x)/dx + √a.d(1/√x)/dx

⇒ 1/√a.1/2√x + √a(-1/2).1/x√x

⇒ 1/2x{√x/a + (-√a/x)}

⇒ 2x.dy/dx = √x/a – √a/x

Multiplying both side by y = √x/a + √a/x, we get

⇒ 2xy.dy/dx = (√x/a – √a/x)(√x/a + √a/x)

⇒ (x/a – a/x)

Hence proved.

### Question 23. Find the rate at which function f(x) = x4 – 2x3 + 3x2 + x + 5 changes with respect to x.

Solution:

Given that, f(x) = x4 – 2x3 + 3x2 + x + 5

Now, differentiate w.r.t. x, we get

df(x)/dx = d(x4 – 2x3 + 3x2 + x + 5) / dx

⇒ 4x3 – 6x2 + 6x + 1.

### Question 24. If y = 2x9/3 – 5x7/7 + 6x3 – x, find dy/dx at x = 1.

Solution:

Given that, y = 2x9/3 – 5x7/7 + 6x3 – x   …..(1)

Find that dy/dx at x = 1

Now, differentiate eq(1) on both side w.r.t. x, we get

dy/dx = d(2x9/3 – 5x7/7 + 6x3 – x) / dx

⇒ 2/3dx9/dx – 5/7dx7/dx + 6dx3/dx – dx/dx

⇒ 2/3.9x8 – 5/7.7x6 + 18x2 – 1

⇒ 6x8 – 5x6 + 18x2 – 1.

Put x = 1

⇒ 6(1)8 – 5(1)6 + 18(1)2 – 1

⇒ 6 – 5 + 18 – 1

⇒ 18

### Question 25. If f(x) = λx2 + μx + 12, f'(4) = 15 and f'(2) = 11, then find λ and μ.

Solution:

Given that, f(x) = λx2 + μx + 12  …..(1)

f'(4) = 15 and f'(2) = 11

Find: the value of λ and μ.

Now, differentiate eq(1) w.r.t. x, we get

f(x) = λx2 +μx + 12

f'(x) = 2λx + μ

Now put f'(4) = 15, we get

⇒ 2λ(4) + μ = 15

⇒ 8λ + μ = 15   ……………(1)

Now put, f'(2) = 11

⇒ 2λ(2) + μ = 11

4λ + μ = 11     …………… (2)

From equation (1) and (2), we get

⇒ 4λ = 4

⇒ λ = 1

Now put value of λ in equation (1), we get

⇒ 8(1) + μ = 15

⇒ μ = 7

Hence, the value of λ = 1 and μ = 7

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### Prove that f'(1) = 100f'(0).

Solution:

Given that, f(x) = x100/100 + x99/99 + ………….. + x2/2 + x + 1

Now, differentiate w.r.t. x, we get

⇒ f'(x) = x99 + x98 + ………… + x + 1 + 0  ……………..(1)

From equation (1),

⇒ f'(1) = 1 + 1 + …………….(100 times)

⇒ 100

Again,

⇒ f'(0) = 0 + 0 + ………….. + 1

⇒ 1

Now,

⇒ f'(1) = 100

⇒ 100 × 1 = 100 × f'(0)

⇒ f'(1) = 100f'(0)

Hence Proved

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