Class 11 RD Sharma Solutions - Chapter 30 Derivatives - Exercise 30.3

In this article, we will delve into the solutions for Exercise 30.3 from Chapter 30 of RD Sharma's Class 11 Mathematics textbook which covers "Derivatives". This chapter is pivotal for understanding how to determine the rate at which a function changes which is fundamental in calculus. The solutions provided will guide us through the steps necessary to solve derivative problems effectively helping to build a strong foundation in the differentiation.

Derivatives

The Derivatives measure how a function's output value changes as its input value changes. In essence, the derivative of a function represents its instantaneous rate of change. It is calculated using various rules and techniques such as the power rule, product rule, and chain rule depending on the form of the function. Understanding derivatives is essential for analyzing the behavior of functions and solving real-world problems involving rates of change.

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) = 3^x + x³ + 3³ with respect to x.

Solution:

Given that, f(x) = 3^x + x³ + 3³

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

⇒ d(3^x + x³ + 3³) / dx

⇒ d(3^x)/dx + d(x³)/dx + d(3³)/dx

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

⇒ 3^x 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) = 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) = (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 + 3log_ex + 2tanx with respect x.

Solution:

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

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

⇒ d( log₃x + 3log_ex + 2tanx)/dx

⇒ 1/log₃ d(logx)/dx + 3.d(log_ex)/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 - 1/2x^3/2) +  (√x + 1/√x) (1 - 1/x²)

⇒ {x/(2√x) - x/(2x^3/2) + 1/2(x^3/2)- 1/(2x^5/2)} + {√x - √x/x² + 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 = a²+ 3a²b + 3ab² + b³]

⇒ 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) = 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 + a₁x^n-1 + a₂x^n-2  + ......... + a_n-1x + a_n) with respect to x.

Solution:

Given that, f(x) = (a₀xⁿ + a₁x^n-1 + a₂x^n-2  + ......... + a_n-1x + a_n)

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

⇒ d(a₀xⁿ + a₁x^n-1 + a₂x^n-2 + ......... + a_n-1x + a_n) / dx

⇒ a₀d(x)ⁿ/dx + a₁d(x)^n-1/dx + a₂d(x)^n-2/dx + .......... + a_n-1d(x)/dx + a_nd(1)/dx

⇒ na₀x^n-1 + (n-1)a₁x^n-2 + .......... + a_n-1 + 0

⇒ na₀x^n-1+ (n-1)a₁x^n-2 + ........... + a_n-1

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

Solution:

Given that, f(x) = 1/sinx + 2^x+3 + 4/logx³ 

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

⇒ d/dx (1/sinx + 2^x+3 + 4/logx³)

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

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

⇒ -cosecx.cotx + 2^x+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/x² 

Question 16. Differentiate f(x) = log(1/√x) + 5x^a - 3a^x + 3√x² + 6(⁴√x^-3) with respect to x.

Solution:

Given that, f(x) = log(1/√x) + 5x^a - 3a^x + 3√x² + 6(⁴√x^-3) 

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

⇒ d/dx {log(1/√x) + 5x^a - 3a^x + ³√x² + 6(⁴√x^-3)}

⇒ d(log(1/√x)/dx + 5d(x^a)/dx - 3(a^x) + d(³√x²)/dx + 6d(⁴√x^-3)/dx

⇒ -1/2.1/x + 5ax^a-1 - 3a^xloga + 2x^-1/3/3 + 6x^-7/4(-3/4)

⇒ -1/2x + 5ax^a-1 - 3a^xloga + 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

​  

Question 26. For the function f(x) = x¹⁰⁰/100 + x⁹⁹/99 + .............. + x²/2 + x + 1.

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

Read More:

• Derivative of a Function
• Derivative Rules