This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Differentiability”.

1. Find the derivative of f(x) = sin(x^{2}).

a) -sin(x^{2})

b) 2xcos(x^{2})

c) -2xcos(x^{2})

d) -2xsin(x^{2})

View Answer

Explanation: Differentiation of the function f(x) = sin(x

^{2}) is done with chain rule. First we differentiate sin function which becomes cos and then differentiate the inner (x

^{2}) which becomes 2x, hence it comes out to be 2xcos(x

^{2}).

2. What is derivative of x^{n}?

a) n

b) nx^{n}

c) nx^{n-1}

d) nx^{n-2}

View Answer

Explanation: It is a standard rule for derivative of a function of this form in which the original power comes in front and the value in the power is decreased by one. Therefore the only option of this form is nx

^{n-1}from the given options.

3. Find derivative of tan(x+4).

a) sec^{2}(x+4)

b) 4 sec^{2}(x+4)

c) 4x sec^{2}(x+4)

d) sec^{2}(x)

View Answer

Explanation: We know that derivative of tanx is sec

^{2}(x), now in the above question we get tan(x+4), hence its derivative comes out to be sec

^{2}(x+4), as the inside expression (x+4) is differentiated into 1.

Therefore the answer is sec

^{2}(x+4).

4. What is value of \(\frac{dy}{dx}\) if x-y = 1?

a) 1

b) 2

c) -1

d) 2

View Answer

Explanation: We know x-y = 1, hence we differentiate it on both sides-:

We get 1- \(\frac{dy}{dx}\) = 0, \(\frac{dy}{dx}\) = 1, hence the value of \(\frac{dy}{dx}\) comes out to be 1.

5. Value after differentiating cos (sinx) is _________

a) sin (sinx).cosx

b) -sin (sinx).cosx

c) sin (sinx)

d) sin (cosx).cosx

View Answer

Explanation: We differentiate the given function with the help of chain rule so we first differentiate the outer function which becomes –sin and then we differentiate the inner function sinx which is differentiated and comes out to be cosx, hence the differentiated function comes out to be -sin (sinx).cosx.

6. Value after differentiating cos (x^{2}+5) is ________

a) 5.sin (x^{2}+5)

b) -sin (x^{2}+5).2x

c) sin (x^{2}+5).2x

d) cos (x^{2}+5).2x

View Answer

Explanation: We differentiate the given function with help of chain rule and hence the outer function becomes –sin and the inner function is differentiated into 2x, therefore the answer comes out to be -sin (x

^{2}+5).2x.

7. Find \(\frac{dy}{dx}\) of 2x+3y = sinx.

a) \(\frac{cosx-2}{3}\)

b) \(\frac{cosx-2}{2}\)

c) \(\frac{cosx-3}{2}\)

d) \(\frac{sinx-2}{3}\)

View Answer

Explanation: Differentiating on both sides we get 2 + 3\(\frac{dy}{dx}\) = cosx.

3\(\frac{dy}{dx}\) = cosx-2.

\(\frac{dy}{dx} = \frac{cosx-2}{3}\).

8. What is derivative of cotx?

a) tanx

b) –sec^{2}x

c) –cosec^{2}x

d) cosec^{2}x

View Answer

Explanation: The derivative of cotx is –cosec

^{2}x, as this function has a fixed derivative like sinx has its derivative cosx. Therefore the answer to the above question is –cosec

^{2}x.

9. If \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2}), \frac{-1}{\sqrt{3}} < x < \frac{-1}{\sqrt{3}}\)

a) 3

b) \(\frac{3}{1+x}\)

c) –\(\frac{3}{1+x^2}\)

d) \(\frac{3}{1+x^2}\)

View Answer

Explanation: Given function is \(y = tan^{-1}(\frac{3x-x^3}{1-3x^2})\),

Now taking RHS and substituting x = tang in it and then we get,

y = \(tan^{-1}(\frac{3tang-tang^3}{1-3tang^2})\), Now it becomes the expansion of the function tan3g,

Hence the given function becomes y= tan

^{-1}(tan3g). Which is equal to 3g, now substituting the value of g= tan

^{-1}x, now after differentiating both sides we get the answer \(\frac{3}{1+x^2}\).

10. Find \(\frac{dy}{dx}\) of y = sin (ax + b).

a) a.cos (ax + b)

b) b.sin (ax + b)

c) a.sin (ax + b)

d) b.cos (ax + b)

View Answer

Explanation: We differentiate the given function with the help of chain rule and hence

the outer function is differentiated into cos, and the inner function comes out to be a and the constant b becomes 0, which is multiplied to the whole function and the answer comes out to be

=> a.cos (ax + b).

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