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The derivative of $\sin^3(2x)$ is
A$3\sin^2(2x)\cos(2x)$
B$6\sin^2(2x)\cos(2x)$
C$6\sin(2x)\cos^2(2x)$
D$3\sin^2x\cos 2x$
Answer & Solution
Correct answer: B. $6\sin^2(2x)\cos(2x)$
Let $u=\sin(2x)$. Then $y=u^3$, so $\dfrac{dy}{dx}=3u^2\dfrac{du}{dx}$. Since $\dfrac{d}{dx}(\sin 2x)=2\cos 2x$, we get $\dfrac{dy}{dx}=3\sin^2(2x)\cdot 2\cos 2x=6\sin^2(2x)\cos 2x$.
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