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  • If x=e^{\cos 2 t} and y=e^{\text {sn } n 2 t}, prove that \frac{{dy}}{{dx}}=\frac{-{y} \log \{x}}{{x} \log {y}}

If \mathrm{x}=e^{\cos 2 t} and \mathrm{y}=\mathrm{e}^{ \sin 2 t}, prove that  \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\mathrm{y} \log \mathrm{x}}{\mathrm{x} \log \mathrm{y}} \text { . }

Answers (1)

\\x = e\textsuperscript{cos2t} and y = e\textsuperscript{sin2t} \\Now x = e\textsuperscript{cos2t},

Taking log on both sides to get,

\\\log x = cos 2t \\$For y$ = e\textsuperscript{sin2t}

Taking log on both sides we get,

log y = sin 2t

\\\therefore cos\textsuperscript{2}2t + sin\textsuperscript{2}2t = (log x)\textsuperscript{2} + (log y)\textsuperscript{2} \\ 1 = (log x)\textsuperscript{2} + (log y)\textsuperscript{2}

Differentiating w.r.t x,

\\\Rightarrow 0=2 \log{x} \frac{1}{x}+2 \log y \frac{1}{y} \\ \\\frac{d y}{d x}=-\frac{y \log x}{x \log y}

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infoexpert22

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