If y^x = e^(y-x), then prove that dy/dx = (1 + logy)^2/log y


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


Given, y^x = e^{y-x}

Taking logrithm both sides, we get log

\log y^x = \log e^{y-x}

\Rightarrow x.\log y = (y-x)\log e

\Rightarrow  x.\log y = (y-x)

\Rightarrow x.\log y + x = y

\Rightarrow x(\log y + 1) = y

\Rightarrow x = \dfrac{y}{1+ \log y}

Differentiating both sides with respect to y, we get

\Rightarrow \dfrac{d x}{dy} = \dfrac{(1 + \log y)\frac{d}{dy}.y-y.\frac{d}{dy}(1+ \log y)}{(1+ \log y)^2}

\Rightarrow \dfrac{d x}{dy} = \dfrac{(1+\log y).1 - y.(0+\frac{1}{y})}{(1+ \log y)^2}

\Rightarrow \dfrac{d x}{dy} = \dfrac{1 + \log y - 1}{(1+ \log y)^2}

\Rightarrow \dfrac{d x}{dy} = \dfrac{\log y}{(1+ \log y)^2}

\Rightarrow \dfrac{d y}{dx} = \dfrac{(1+ \log y)^2}{\log y}

Some other question

Q 1 :If tan inverse y/x = log root x^2 + y^2, then prove that

Q 2: Find the value of \dfrac{dy}{dx} at \theta = \dfrac{\pi}{4}, if x = ae^{\theta}(\sin \theta - \cos \theta) and y = ae^{\theta}(\sin \theta + \cos \theta).          ……..   [CBSC 2008, 2014]

Team Gmath

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