Question:

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

Solution:

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

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

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

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

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

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

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]