What is the derivative of 2^{(sinx)²} with respect to sin x?

Correct Answer is : sinx2(sinx)2ln4sinx 2^{(sinx)² }ln 4sinx2(sinx)2ln4

Correct Answer is : sinx2(sinx)2ln4sinx 2^{(sinx)² }ln 4sinx2(sinx)2ln4

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UPSC NDA 2 2019 Mathematics Solved Paper

Question : 97 of 120

Marks: +1, -0

2^{(sinx)^{2}}

with respect to sin x?

Solution:

Concept:

Derivative of

a^{u}

with respect to

u

a^{u} \ln a

, and we apply the chain rule when differentiating with respect to another variable.

Explanation:

We need

\frac{d}{d(\sin x)} \left( 2^{(\sin x)^{2}} \right)

Let

t = \sin x

. Then the expression becomes

2^{t^{2}}

Using

\frac{d}{dt} (a^{u}) = a^{u} \ln a \cdot \frac{du}{dt}

\frac{d}{dt} \left( 2^{t^{2}} \right) = 2^{t^{2}} \cdot \ln 2 \cdot \frac{d(t^{2})}{dt} = 2^{t^{2}} \cdot \ln 2 \cdot 2t

Substitute

t = \sin x

2^{(\sin x)^{2}} \cdot \ln 2 \cdot 2 \sin x = 2^{(\sin x)^{2}} \cdot (2 \ln 2) \cdot \sin x

Since

2 \ln 2 = \ln (2^{2}) = \ln 4

, the result simplifies to

\sin x \cdot 2^{(\sin x)^{2}} \cdot \ln 4

Answer:

\sin x \cdot 2^{(\sin x)^{2}} \ln 4

(Option A).

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