If y = u² + log u and u = e^x, then find \(\rm \frac{dy}{dx}:\)

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Question

This question was previously asked in

AAI ATC Junior Executive 21 Feb 2023 Shift 1 Official Paper

Answer 1

Given:

y = u2 + log u and u = ex

Concept:

Use concept of chain rule

\(\rm \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\)

And \(\rm \frac{d}{dx}e^x=e^x\)

Calculation:

\(\rm u=e^x\)

Now, differentiate with respect to \(\rm x\)

\(\rm \frac{du}{dx}=e^x\)

\(\rm y=u^2+log\ u\)

Now differentiate with respect to \(\rm x\)

\(\rm \frac{dy}{dx}=2u \frac{du}{dx}+\frac{1}{u} \frac{du}{dx}\)

Now put the values of \(\rm u \ and \ \frac{du}{dx}\)

\(\rm \frac{dy}{dx}=2e^x\cdot \ e^x+\frac{1}{e^x} \ e^x\)

\(\rm \implies \frac{dy}{dx}=2e^{2x}+1\)

Hence the option (3) is correct.

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