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

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(D) Given,$y = \sin^{n} x \cos nx$. Applying the product rule $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$: $\frac{dy}{dx} = \sin^{n} x \fra

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

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(D) Given,$y = \sin^{n} x \cos nx$. Applying 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 $\cos(A+B) = \cos A \cos B - \sin A \sin B$: $\frac{dy}{dx} = n \sin^{n-1} x \cos (nx + x)$ $\frac{dy}{dx} = n \sin^{n-1} x \cos (n+1) x$

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

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