If $y = {\sin ^n}x\cos nx,$ then $\frac{{dy}}{{dx}}$ is

$n{\sin ^{n - 1}}x\sin \left( {n + 1} \right)x$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y = {\sin ^n}x\cos nx,

then

\frac{{dy}}{{dx}}

View answer and explanation

Correct option: D

n{\sin ^{n - 1}}x\cos \left( {n + 1} \right)x

The correct answer is

n{\sin ^{n - 1}}x\cos \left( {n + 1} \right)x

because it results from applying the product rule and trigonometric identities to the given function.

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