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UPSC NDA I 2018 Mathematics Solved Paper

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Question : 81 of 120

Marks: +1, -0

If

y=e^{x^{2}}sin2x

\frac{dy}{dx}

at x = π equal to?

$(1+\pi )e^{\pi ^{2}}$
$2\pi e^{\pi ^{2}}$
$2e^{\pi ^{2}}$
$e^{\pi ^{2}}$

Solution:

Concept:

Use the product rule of differentiation:

\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)

.

Explanation:

Given

y = e^{x^2} \sin 2x

Differentiate

y

with respect to

x

:

\frac{dy}{dx} = \frac{d}{dx}(e^{x^2}) \cdot \sin 2x + e^{x^2} \cdot \frac{d}{dx}(\sin 2x)

\frac{d}{dx}(e^{x^2}) = e^{x^2} \cdot 2x

(chain rule)

\frac{d}{dx}(\sin 2x) = 2 \cos 2x

Thus,

\frac{dy}{dx} = 2x e^{x^2} \sin 2x + e^{x^2} \cdot 2 \cos 2x = 2e^{x^2}(x \sin 2x + \cos 2x)

Now evaluate at

x = \pi

:

\sin(2\pi) = 0

,

\cos(2\pi) = 1

So

\left. \frac{dy}{dx} \right|_{x=\pi} = 2e^{\pi^2} (\pi \cdot 0 + 1) = 2e^{\pi^2}

Answer:

2e^{\pi^2}

which matches option C.

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