Q.1 (d) Find n^th derivative of y=e^axsin²x .

y=e^ax·1-cos2x/2=1/2e^ax-1/2e^axcos2x . Standard results: Dⁿ(e^ax)=aⁿe^ax ; Dⁿ[e^axcos(bx+c)]=rⁿe^axcos(bx+c+nφ) where r=√(a²+b²),\ φ=tan⁻¹b/a . Here b=2 : r=√(a²+4),\ φ=tan⁻¹2/a . y_n=1/2[aⁿe^ax-(a²+4)^(n/2)e^axcos(2x+ntan⁻¹2/a)]

Find \(n^{th}\) derivative of \(y=e^{ax}\sin^2x\).

Q.1 (d) Find \(n^{th}\) derivative of \(y=e^{ax}\sin^2x\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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\(y=e^{ax}\cdot\frac{1-\cos2x}{2}=\frac12e^{ax}-\frac12e^{ax}\cos2x\).

Standard results: \(D^n(e^{ax})=a^ne^{ax}\); \(D^n\left[e^{ax}\cos(bx+c)\right]=r^ne^{ax}\cos(bx+c+n\phi)\) where \(r=\sqrt{a^2+b^2},\ \phi=\tan^{-1}\frac ba\).

Here \(b=2\): \(r=\sqrt{a^2+4},\ \phi=\tan^{-1}\frac2a\).

\(y_n=\frac12\left[a^ne^{ax}-(a^2+4)^{n/2}e^{ax}\cos\left(2x+n\tan^{-1}\frac{2}{a}\right)\right]\)

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

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Find \(n^{th}\) derivative of \(y=e^{ax}\sin^2x\).

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