TEST ON DIFFERENTIATION FOR CA FOUNDATION

Problem 1 :

If f(x) = x^k and f^’(1) = 10, then the value of k is :

(a) 10     (b) -10     (c) 1/10      (d) None

Solution :

f(x) = x^k

f´(x) = kx^{k – 1}

Given, f´(1) = 10

f´(1) = k(1)^{k – 1}

f´(1) = k

f´(1) = k

k = 10

So, the value of k is10.

So, option (a) is correct.

Problem 2 :

If y = 4x³ – 7x⁴ then dy/dx is

(a)  2x(14x² – 6x)     (b) 2x(-14x² + 6x)

(c) 2x(14x² + 6x)       (d) None

Solution :

y = 4x³ – 7x⁴

Differentiation with respect to ‘x’.

dy/dx = 4(3x²) - 7(4x³)

dy/dx = 12x² - 28x³

dy/dx = 2x(6x - 14x²)

So, option (b) is correct.

Problem 3 :

If x² + y² = a², find dy/dx.

(a)  y/x        (b) – y/x       (c) –x/y       (d) x/y

Solution :

x² + y² = a²

Differentiating with respect to x, we get

2x + 2y(dy/dx) = 0

dy/dx = -2x/2y

= -x/y

So, option (c) is correct.

Problem 4 :

Let x = at³, y = a/t³. Then dy/dx =

(a) -1/t⁶      (b) -3a/t⁶     (c) 1/3at⁶     (d) None

Solution :

dy/dt × dt/dx = -3a/t⁴ × 1/3at²

dy/dx = -1/t⁶

So, option (a) is correct.

Problem 5 :

If y = e^3x, find y’’.

(a) 6e^3x        (b) 3e^3x        (c) 12e^3x        (d) 9e^3x

Solution :

y = e^3x

y´ = 3 ⋅ e^3x

y´´ = 9 ⋅ e^3x

So, option (d) is correct.

Problem 6 :

If x^m yⁿ = (x + y)^{m + n}, then find dy/dx :

(a)x/y      (b) y/x      (c) xy     (d) None

Solution :

x^m ⋅ yⁿ = (x + y)^{m + n}

taking log on both sides.

log(x^m ⋅ yⁿ) = log(x + y)^{m + n}

logx^m + logyⁿ = (m + n) ⋅ log(x + y)

m logx + n logy = (m + n) ⋅ log(x + y)

Differentiating both sides with respect to ‘x’.

m/x + n/y ⋅ dy/dx = (m + n)/(x + y) ⋅ (1 + dy/dx)

So, option (b) is correct.

Problem 7 :

If y = log X^x then dy/dx is equal to :

(a) log ex (b) log e/x (c) log x/e (d) 1

Solution :

Given, y = log X^x

= x log x

Differentiate with respect to ‘x’.

Using product rule

u = x and v = log x

du = dx and dv = 1/x

Product rule :

d(uv) = udv + vdu

dy/dx = x(1/x) + logx

= 1 + log x

= log e + log x

= log(ex)

So, option (a) is correct.

Problem 8 :

If y = (x^1/3 - x^-1/3)³, then dy/dx is

(a) 1 + x^-2 + x^-2/3 – x^-4/3       (b) 1 + x^-2 + x^-2/3 – x^-4/3

(c) 1 – x^-2 + x^-2/3 – x^-4/3       (d) None of these

Solution :

y = (x^1/3 - x^-1/3)³

(a – b)³ = a³ – 3a²b + 3b²a – b³

y = x – 3x^1/3 + 3x^-1/3 – x^-1

dy/dx = 1 – x^-2/3 – x^-4/3 + x^-2

So, option (a) is correct.

Problem 9 :

If f(x) = e^{ax^2 + bx + c} the f’(x) is

(a) e^{ax^2 + bx + c}     (b) e^{ax^2 + bx + c} (2ax + b)

(c) 2ax + b               (d) None of these

Solution :

f(x) = e^{ax^2 + bx + c}

f´(x) = (2ax + b) e^{ax^2 + bx + c}

So, option (b) is correct.

Problem 10 :

If u = 3t⁴ + 5t³ + 2t² + t + 4, then the value of du/dt at t = -1 is

(a) 0      (b) 1       (c) 2        (d) 5

Solution :

u = 3t⁴ + 5t³ + 2t² + t + 4

du/dt = 12t³ + 15t² + 4t + 1

Given, t = -1

= 12(-1)³ + 15(-1)² + 4(-1) + 1

= -12 + 15 - 4 + 1

= 0

So, option (a) is correct.

Problem 11 :

If y = (x – 1) (x + 1), find d²y/dx².

(a) 2       (b) -1       (c) 0          (d) x²

Solution :

y = (x – 1) (x + 1)

= x² + x – x – 1

= x² - 1

dy/dx = 2x

d²y/dx² = 2

So, option (a) is correct.

Problem 12 :

The gradient of the curve

y = 2x³ – 3x² – 12x + 8 at x = 0 is

(a) -12      (b) 12      (c) 0     (d) None of these        

Solution :

y = 2x³ – 3x² – 12x + 8

dy/dx = 2 ⋅ 3x² – 3 ⋅ 2x - 12

dy/dx = 6x² – 6x – 12

substitute x = 0

= 6(0)² – 6(0) – 12

= -12

So, option (a) is correct.

Problem 13 :

The gradient of the curve

y = 2x³ – 5x² – 3x at x = 0 is

(a) 3      (b) -3     (c) 1/3      (d) None of these          

Solution :

y = 2x³ – 5x² – 3x

dy/dx = 2 ⋅ 3x² – 5 ⋅ 2x - 3

dy/dx = 6x² – 10x – 3

substitute x = 0

= 6(0)² – 10(0) – 3

= -3

So, option (b) is correct.

Problem 14 :

The derivative of y = √(x + 1) is

(a) 1/√(x + 1)       (b) -1/√(x + 1)     (c) 1/2√(x + 1)

Solution :

y = √(x + 1)

dy/dx = 1/2(x + 1)^-1/2

dy/dx = 1/2√(x + 1)

So, option (c) is correct.

Problem 15 :

If f(x) = (x² + 1)/(x² – 1) then f’(x) is

(a) -4x/(x² – 1)²      (b) 4x/(x² – 1)²

(c) x/(x² – 1)²         (d) None of these  

Solution :

f(x) = (x² + 1)/(x² – 1)

Using quotient rule,

u = x² + 1 and v = x² - 1

du = 2x and dv = 2x

d(u/v) = (vdu - udv) / v²

So, option (a) is correct.