Q.3 (A) (ii) If x^y=e^(x-y) then prove that dy/dx=log x/(1+log x)² .

Answer: Proved below. Taking log on both sides: ylog x=(x-y)log e=x-y . therefore ylog x+y=x⇒ y(1+log x)=x⇒ y=x/1+log x . Differentiating w.r.t. x by the quotient rule: dy/dx=(1+log x)(1)-x·1/x/(1+log x)²=1+log x-1/(1+log x)²=log x/(1+log x)² .

If \(x^y=e^{x-y}\) then prove that \(\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}\).

Q.3 (A) (ii) If \(x^y=e^{x-y}\) then prove that \(\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}\).

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: Proved below.

Taking log on both sides: \(y\log x=(x-y)\log e=x-y\).

\(\therefore y\log x+y=x\Rightarrow y(1+\log x)=x\Rightarrow y=\frac{x}{1+\log x}\).

Differentiating w.r.t. \(x\) by the quotient rule:

\(\frac{dy}{dx}=\frac{(1+\log x)(1)-x\cdot\frac{1}{x}}{(1+\log x)^2}=\frac{1+\log x-1}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}\).

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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If \(x^y=e^{x-y}\) then prove that \(\frac{dy}{dx}=\frac{\log x}{(1+\log x)^2}\).

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