Find \(n^{th}\) derivative of \(y=\sin2x\cos6x\).

Q.1 (c) Find \(n^{th}\) derivative of \(y=\sin2x\cos6x\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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Best answer

Using \(2\sin A\cos B=\sin(A+B)+\sin(A-B)\): \(y=\frac12[\sin8x+\sin(-4x)]=\frac12[\sin8x-\sin4x]\).

Using \(D^n\sin(ax)=a^n\sin\left(ax+\frac{n\pi}2\right)\):

\(y_n=\frac12\left[8^n\sin\left(8x+\frac{n\pi}2\right)-4^n\sin\left(4x+\frac{n\pi}2\right)\right]\)

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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