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

### Question 1. Differentiate f(x) = x^{4} – 2sinx + 3cosx with respect to x.

**Solution:**

Given that, f(x) = x

^{4}– 2sinx + 3cosxNow, differentiate w.r.t. x, we get

⇒ d(x

^{4 }– 2sinx + 3cosx) / dx⇒ d(x

^{4})/dx – 2.d(sinx)/dx + 3.d(cosx)/dx⇒ 4x

^{3}– 2cosx – 3sinx.

### Question 2. Differentiate f(x) = 3^{x} + x^{3} + 3^{3} with respect to x.

**Solution:**

Given that, f(x) = 3

^{x}+ x^{3}+ 3^{3}Now, differentiate w.r.t. x, we get

⇒

d(3^{x}+ x^{3}+ 3^{3}) / dx⇒ d(3

^{x})/dx + d(x^{3})/dx + d(3^{3})/dx⇒ 3

^{x}log3 + 3x^{2}+ 0 [As we know, d(a^{x})/dx = a^{x}loga]⇒ 3

^{x}log3 + 3x^{2}

### Question 3. Differentiate f(x) = x^{3}/3 – 2√x + 5/x^{2} with respect to x.

**Solution:**

Given that, f(x) = x

^{3}/3 – 2√x + 5/x^{2}Now, differentiate w.r.t. x, we get

⇒

d(x^{3}/3 – 2√x + 5/x^{2}) / dx⇒ 1.d(x

^{3})/3dx – 2d(√x)/dx + 5d(x^{-2})/dx⇒ 1/3.3x

^{2}– 2.1/2.1/√x + 5(-2) x^{-3}⇒ x

^{2 }– x^{-1/2 }– 10x^{-3}⇒ x

^{2 }– 1/√x – 10/x^{3}

### Question 4. Differentiate f(x) = e^{xloga} + e^{alogx} + e^{aloga} with respect to x.

**Solution:**

Given that, f(x) = e

^{xloga}+ e^{alogx}+ e^{aloga}Now, differentiate w.r.t. x, we get

⇒ d(e

^{xloga}+ e^{alogx}+ e^{aloga})⇒ d(e

^{xloga})/dx + d(e^{alogx})/dx + d(e^{aloga})/dx⇒ e

^{xloga}.loga + e^{alogx}.a/x + 0 [As we know, e^{aloga}is constant]⇒ loga.e

^{xloga}+ a/x.e^{alogx}⇒ loga.a

^{x}+ a/x x^{a }[Here, a^{x}can be written as a e^{xloga}]⇒ a

^{x}loga + ax^{a-1}

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

**Solution:**

Given that, f(x) = (2x

^{2}+ 1)(3x + 2)Now, differentiate w.r.t. x, we get

⇒

d(2x^{2}+ 1)(3x + 2)/dx⇒ (3x + 2)d(2x

^{2}+ 1)/dx + (2x^{2}+ 1)d(3x + 2)/dx⇒ (3x + 2)(4x+0) + (2x

^{2}+ 1)(3+ 0)⇒ (12x

^{2}+ 8x + 6x^{2}+ 3)⇒ 18x

^{2}+ 8x + 3.

### Question 6. Differentiate f(x) = log_{3}x + 3log_{e}x + 2tanx with respect x.

**Solution:**

Given that, f(x) = log

_{3}x + 3log_{e}x + 2tanxNow, differentiate w.r.t. x, we get

⇒

d(log_{3}x + 3log_{e}x + 2tanx)/dx⇒ 1/log

_{3}d(logx)/dx + 3.d(log_{e}x)/dx + 2.d(tanx)/dx⇒ 1/log

_{3 }× 1/x + 3/x + 2sec^{2}x⇒ 1/xlog

_{3}+ 3/x + 2sec^{2}x

### 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 – 1/2x

^{3/2}) + (√x + 1/√x) (1 – 1/x^{2})⇒ {x/(2√x) – x/(2x

^{3/2}) + 1/2(x^{3/2})- 1/(2x^{5/2})} + {√x – √x/x^{2}+ 1/√x – 1/x^{5/2}}⇒ (1.√x/2 – 1/2√x + 1/2x

^{3/2}– 1/2x^{5/2}+ √x – 1/x^{3/2}+ 1/√x – 1/x^{5/2})⇒ (3√x/2 + √x/2 – 1/2x

^{3/2}– 3/2x^{5/2})⇒ 3x

^{1/2}/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(x

^{3/2}+ 3x.1/x + 3√x.1/x + 1/x^{3/2})/dx [As we know that, (a + b)^{3}= a^{2}+ 3a^{2}b + 3ab^{2}+ b^{3}]⇒ d(x

^{3/2}+ 3x^{1/2}+ 3x^{-1/2}+ x^{-3/2})/dx⇒ 3x

^{1/2}/2 + 3x^{-1/2}/2 + 3.(-1/2).x^{-3/2}– 3x^{-5/2}/2⇒ 3x

^{1/2}/2 – 3x^{-5/2}/2 + 3x^{-1/2}/2 – 3x^{-3/2}/2.

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

**Solution:**

Given that, f(x) = 2x

^{2}+ 3x + 4 /xNow, differentiate w.r.t. x, we get

⇒ d(2x

^{2}+ 3x + 4 /x) / dx⇒d(2x

^{2}/x + 3x/x + 4/x) /dx⇒ d(2x + 3 + 4x

^{-1}) / dx⇒ 2- 4/x

^{2}

### Question 10. Differentiate f(x) = (x^{3}+ 1) (x – 2) / x^{2 }with respect x.

**Solution:**

Given that, f(x) = (x

^{3}+ 1) (x – 2) / x^{2}Now, differentiate w.r.t. x, we get

⇒ d{(x

^{3}+ 1) (x – 2) / x^{2}} / dx⇒ d{(x

^{4}– 2x^{3}+x – 2)/ x^{2}} / dx⇒ d(x

^{2}– 2x + x^{-1}– 2x^{-2}) / dx⇒ d(x

^{2})/dx – 2d(x)/dx + d(x^{-1})/dx – 2d(x^{-2})/dx⇒ 2x – 2 – 1/x

^{2}+ 4/x^{3}⇒ 2x – 2 – 1/x

^{2}+ 4/x^{3}

### 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(-cosec

^{2}x) + 0 + c(-cosecx.cotx)⇒ -acosec

^{2}x – 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 – 3cosec

^{2}x – 4sec^{2}x

### Question 13. Differentiate f(x) = (a_{0}x^{n} + a_{1}x^{n-1} + a_{2}x^{n-2 }+ ……… + a_{n-1}x + a_{n}) with respect to x.

**Solution:**

Given that, f(x) = (a

_{0}x^{n}+ a_{1}x^{n-1}+ a_{2}x^{n-2 }+ ……… + a_{n-1}x + a_{n})Now, differentiate w.r.t. x, we get

⇒ d(a

_{0}x^{n}+ a_{1}x^{n-1}+ a_{2}x^{n-2}+ ……… + a_{n-1}x + a_{n}) / dx⇒ a

_{0}d(x)^{n}/dx + a_{1}d(x)^{n-1}/dx + a_{2}d(x)^{n-2}/dx + ………. + a_{n-1}d(x)/dx + a_{n}d(1)/dx⇒ na

_{0}x^{n-1}+ (n-1)a_{1}x^{n-2}+ ………. + a_{n-1}+ 0⇒ na

_{0}x^{n-1}+ (n-1)a_{1}x^{n-2}+ ……….. + a_{n-1}

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

**Solution:**

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

^{x+3}+ 4/logx^{3 }Now, differentiate w.r.t. x, we get

⇒ d/dx (1/sinx + 2^{x+3}+ 4/logx^{3})⇒ d(cosecx)/dx + 2

^{3}d(2^{x})/dx + 4/log3 × d(logx)/dx [As we know that, log_{b}a = loga/logb]⇒ -cosecx.cotx + 8 × 2log2 + 4/log3 × 1/x [Since, d(a

^{x})/dx = a^{x}loga]⇒ -cosecx.cotx + 2

^{x+3}log2 + 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)(2x

^{2 }– 1)/x}⇒ d/dx (2x

^{3}+ 10x^{2}– x – 5 / x)⇒ d(2x

^{2}+ 10x – 1 – 5x^{-1})/dx⇒ 2d(x

^{2})/dx + 10d(x)/dx – d(1)/dx – 5d(x^{-1})/dx⇒ 2.2x + 10 – 0 + 5/x

^{2}⇒ 4x + 10 + 5/x

^{2 }

### Question 16. Differentiate f(x) = log(1/√x) + 5x^{a} – 3a^{x} + 3√x^{2} + 6(^{4}√x^{-3}) with respect to x.

**Solution:**

Given that, f(x) = log(1/√x) + 5x

^{a}– 3a^{x}+ 3√x^{2}+ 6(^{4}√x^{-3})Now, differentiate w.r.t. x, we get

⇒ d/dx {log(1/√x) + 5x

^{a}– 3a^{x}+^{3}√x^{2}+ 6(^{4}√x^{-3})}⇒ d(log(1/√x)/dx + 5d(x

^{a})/dx – 3(a^{x}) + d(^{3}√x^{2})/dx + 6d(^{4}√x^{-3})/dx⇒ -1/2.1/x + 5ax

^{a-1}– 3a^{x}loga + 2x^{-1/3}/3 + 6x^{-7/4}(-3/4)⇒ -1/2x + 5ax

^{a-1}– 3a^{x}loga + 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.cosec

^{2}x + 0⇒ -cosec

^{2}x.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{sin

^{2}(x/2) + cos^{2}(x/2) + 2sin(x/2).cos(x/2)}/dx⇒ d(1 + sinx)/dx [As we know that sin

^{2}x + cos^{2}x = 1]⇒ 0 + cosx [As we know that sin

^{2}x= 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 + 3cosec

^{2}xNow put x = π/4

⇒ -2cosec(π/4).cot(π/4) + 3cosec

^{2}(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) = 2x^{6} + x^{4} – 1 at x = 1.

**Solution:**

Given that f(x) = 2x

^{6}+ x^{4}– 1 at x = 1.Find the slope of the tangent at a point x = 1

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

⇒ d(2x

^{6}+ x^{4}-1)/dx⇒ 2dx

^{6}/dx + dx^{4}/dx – d.1/dx⇒ 12x

^{5}+ 4x^{3}– 0⇒ 12x

^{5}+ 4x^{3}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) = x^{4} – 2x^{3} + 3x^{2} + x + 5 changes with respect to x.

**Solution:**

Given that, f(x) = x

^{4}– 2x^{3}+ 3x^{2}+ x + 5Now, differentiate w.r.t. x, we get

df(x)/dx = d(x^{4}– 2x^{3}+ 3x^{2}+ x + 5) / dx⇒ 4x

^{3}– 6x^{2}+ 6x + 1.

### Question 24. If y = 2x^{9}/3 – 5x^{7}/7 + 6x^{3} – x, find dy/dx at x = 1.

**Solution:**

Given that, y = 2x

^{9}/3 – 5x^{7}/7 + 6x^{3}– 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(2x^{9}/3 – 5x^{7}/7 + 6x^{3}– x) / dx⇒ 2/3dx

^{9}/dx – 5/7dx^{7}/dx + 6dx^{3}/dx – dx/dx⇒ 2/3.9x

^{8}– 5/7.7x^{6}+ 18x^{2}– 1⇒ 6x

^{8}– 5x^{6}+ 18x^{2}– 1.Put x = 1

⇒ 6(1)

^{8}– 5(1)^{6}+ 18(1)^{2}– 1⇒ 6 – 5 + 18 – 1

⇒ 18

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

**Solution:**

Given that, f(x) = λx

^{2}+ 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) = λx

^{2}+μx + 12f'(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

### Question 26. For the function f(x) = x^{100}/100 + x^{99}/99 + ………….. + x^{2}/2 + x + 1.

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

**Solution:**

Given that, f(x) = x

^{100}/100 + x^{99}/99 + ………….. + x^{2}/2 + x + 1Now, differentiate w.r.t. x, we get

⇒ f'(x) = x

^{99}+ x^{98}+ ………… + 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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