frac{d}{dx}[sin^n x cos nx] = | vedclass

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(A) Let $y = \sin^n x \cos nx$. Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^n x \frac{d}{dx}(

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

Vedclass · Answer

(A) Let $y = \sin^n x \cos nx$. Using the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^n x \frac{d}{dx}(\cos nx) + \cos nx \frac{d}{dx}(\sin^n x)$ $\frac{dy}{dx} = \sin^n x (-n \sin nx) + \cos nx (n \sin^{n-1} x \cos x)$ $\frac{dy}{dx} = n \sin^{n-1} x [\cos x \cos nx - \sin x \sin nx]$ Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $\frac{dy}{dx} = n \sin^{n-1} x \cos(nx + x) = n \sin^{n-1} x \cos(n+1)x$.

$\frac{d}{dx}[\sin^n x \cos nx] = $

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