Differentiation of Implicit Functions MCQ Quiz - Objective Question with Answer for Differentiation of Implicit Functions - Download Free PDF

Last updated on Jul 1, 2025

Latest Differentiation of Implicit Functions MCQ Objective Questions

Differentiation of Implicit Functions Question 1:

Comprehension:

Consider the following for the two (02) items that follow: 

 Let (x+y)p+q=xpyq, where p,q are positive integers.

If p+q=10, then what is  equal to?

Answer (Detailed Solution Below)

Option 1 :

Differentiation of Implicit Functions Question 1 Detailed Solution

Calculation:

Given,

and .

Differentiate both sides with respect to implicitly:

Left side:

Right side (product rule):

Rearrange to collect terms and use

.

After cancellation of the common factor , you obtain:

∴ 

Hence, the correct answer is Option 1. 

Differentiation of Implicit Functions Question 2:

Comprehension:

Consider the following for the two (02) items that follow: 

 Let (x+y)p+q=xpyq, where p,q are positive integers.

The derivative of y with respect to x

  1. depends on p only
  2. depends on q only
  3. depends on both p and qc
  4. is independent of both p and q

Answer (Detailed Solution Below)

Option 4 : is independent of both p and q

Differentiation of Implicit Functions Question 2 Detailed Solution

Calculation:

Given,

Differentiate implicitly w.r.t. :

Rearrange to collect :

Use to simplify:

∴ , independent of and .

Hence, the correct answer is Option 4.

Differentiation of Implicit Functions Question 3:

 Let (x + y)p + q = xpyp, where p, q are positive integers. If p+q=10, then what is equal to?

Answer (Detailed Solution Below)

Option 1 :

Differentiation of Implicit Functions Question 3 Detailed Solution

Calculation:

Given,

and .

Differentiate both sides with respect to implicitly:

Left side:

Right side (product rule):

Rearrange to collect terms and use

.

After cancellation of the common factor , you obtain:

∴ 

Hence, the correct answer is Option 1. 

Differentiation of Implicit Functions Question 4:

 Consider the following for the two (02) items that follow: 

 Let (x+y)p+q=xpyq, where p,q are positive integers.

The derivative of y with respect to x

  1. depends on p only
  2. depends on q only
  3. depends on both p and q
  4. is independent of both p and q

Answer (Detailed Solution Below)

Option 4 : is independent of both p and q

Differentiation of Implicit Functions Question 4 Detailed Solution

Calculation:

Given,

Differentiate implicitly w.r.t. :

Rearrange to collect :

Use to simplify:

∴ , independent of and .

Hence, the correct answer is Option 4.

Differentiation of Implicit Functions Question 5:

If 2x + 2y = 2x+y, then

  1. 1 - 2y
  2. 1 - 2-y
  3. 1 + 2y
  4. 1 + 2-y
  5. 2y

Answer (Detailed Solution Below)

Option 1 : 1 - 2y

Differentiation of Implicit Functions Question 5 Detailed Solution

Concept:

Calculation:

Given 2x + 2y = 2x+y

We know that 2a+b = 2a⋅ 2b

⇒ 2x + 2y = 2x ⋅ 2y

⇒ 

⇒ 2-y + 2-x = 1  ..(1)

Differentiating the above equation with respect to x:

⇒ (-2- y - 2- x) log 2 = 0

⇒ 

⇒ 

⇒  = - 2y + 1

⇒  = 1 - 2y

The required value of   is 1 - 2y .  

Top Differentiation of Implicit Functions MCQ Objective Questions

Answer (Detailed Solution Below)

Option 2 :

Differentiation of Implicit Functions Question 6 Detailed Solution

Download Solution PDF

Calculation:

xe = e 

Taking log on both sides, we get

⇒ log xe = log e 

⇒ e log x = (x2 + y2) log e             

[∵ log mn = n log m]

⇒ e log x = x2 + y2                       

[∵ log e = 1]

Differentiating w.r.t x, we get

⇒ e() = 

⇒ 

∴ 

Answer (Detailed Solution Below)

Option 3 :

Differentiation of Implicit Functions Question 7 Detailed Solution

Download Solution PDF

Concept:

Calculus:

  • .
  •  

Chain Rule of Derivatives:

  • .


Calculation:

Given expression is:

3x + 3y = 3x + y.

Differentiating w.r.t. x and using the chain rule of derivatives, we get:

⇒ 3x (log 3) + 3y (log 3)  = 3x + y (log 3) (1 + )

⇒ 3x + 3y  = 3x + y + 3x + y 

⇒ 

Answer (Detailed Solution Below)

Option 3 :

Differentiation of Implicit Functions Question 8 Detailed Solution

Download Solution PDF

Calculation:

Here, 

Differentiating w.r.t. x, we get

Hence, option (3) is correct.

Answer (Detailed Solution Below)

Option 2 :

Differentiation of Implicit Functions Question 9 Detailed Solution

Download Solution PDF

Calculation:

y + sin-1 (1 - x2) = ex 

y = ex - sin-1 (1 - x2)

Differentiating w.r.t x, we get

Answer (Detailed Solution Below)

Option 2 :

Differentiation of Implicit Functions Question 10 Detailed Solution

Download Solution PDF

Concept:

 

Calculation:

Given: 

Differentiating with respect to x, we get

Answer (Detailed Solution Below)

Option 3 :

Differentiation of Implicit Functions Question 11 Detailed Solution

Download Solution PDF

Concept:

 

Calculation:

Given: y2 = 4ax

Differentiating with respect to x, we get

If y = 3e2x + 2e3x, then  - 5  + 6y equals

  1. e2x + e3y
  2. 6(3e2x + 2e3x)
  3. 1
  4. 0

Answer (Detailed Solution Below)

Option 4 : 0

Differentiation of Implicit Functions Question 12 Detailed Solution

Download Solution PDF

Given

y = 3e2x + 2e3x

Formula used

d(xn)/dx = nxn-1

Solution

⇒dy/dx = 3e2x(2) + 2e3x(3)

⇒dy/dx = 6(e2x + e3x)

⇒d2y/dx= 6(2e2x+3e3x)

As asked in the question,

⇒   - 5  + 6y =

⇒ 12e2x + 18e3x − 30e2x − 30e3x + 18e2x + 12e3x

⇒ 0.

The correct option is 4.

What is the value of  , if y2 + x2 + 3x + 5 = 0 at (0, -3)?

  1. 1
  2. 1.5
  3. 2
  4. 0.5

Answer (Detailed Solution Below)

Option 4 : 0.5

Differentiation of Implicit Functions Question 13 Detailed Solution

Download Solution PDF

Concept:

Chain Rule (Differentiation by substitution): If y is a function of u and u is a function of x

 

Calculation:

Given y2 + x2 + 3x + 5 = 0

Differentiating with respect to x, we get

2y  + 2x +3(1) + 0 = 0

2y + 2x + 3 = 0 

2y  = -(2x + 3)

Now at (0, -3)

 = 0.5

The second derivative of the function y = f(x) given by the equation y2 = 2x is

  1. 1/y3
  2. – 1/y3
  3. 1/y2
  4. – 1/y2

Answer (Detailed Solution Below)

Option 2 : – 1/y3

Differentiation of Implicit Functions Question 14 Detailed Solution

Download Solution PDF

Calculation:

Given function is y2 = 2x;

Differentiating with respect to x on both sides,

2yy' = 2x ⇒ yy’ = 1   - (1)

⇒ y’ = 1/y;

Differentiating (1) again with respect to x on both sides,

⇒ y . y” + y’. y’ = 0   - (2)

If y = cos2 x2, find 

  1. 4x2 sin x2 cos x2
  2. -4x cos x2 sin x2
  3. 2x sin x2 cos x2
  4. -2x cos x2 sin x2

Answer (Detailed Solution Below)

Option 2 : -4x cos x2 sin x2

Differentiation of Implicit Functions Question 15 Detailed Solution

Download Solution PDF

Concept:

cos2x = 2cos2x - 1

sin2x = 2sin x cos x

 

Calculation:

Here, y = cos2 x2

Let, x2 = t 

Differentiating with respect to x, we get

⇒2xdx = dt 

⇒ dt/dx = 2x ....(1)

y = cos2t

=

= - sin2x2 × 2x

= -4x cos x2 sin x2

Hence, option (2) is correct. 

Hot Links: teen patti gold real cash teen patti master online teen patti gold apk download teen patti rules teen patti real cash 2024