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If $y^{\large x}=e^{\large y-x},prove\;that\;\large {\frac{dy}{dx}=\frac{(1+logy)^2}{log y}}$

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Toolbox:
  • A function $f(x,y)$ is said to be implicit if it is jumbled in such a way,that it is not possible to write $y$ exclusively s a function of $x$.
  • $\large\frac{d}{dx}$$\phi(y)=\large\frac{d}{dy}$$\phi(y).\large\frac{dy}{dx}$
Step 1:
$y^{\large x}=e^{\large y-x}$
Take $\log$ on both sides,
$x.\log y=(y-x)\log e$-------(1)
But $\log e=1$
Differentiating w.r.t $x$ on both sides
(Apply product rule to differentiate $x\log y$)
$x.\large\frac{1}{y}\frac{dy}{dx}+$$\log y.1=\large\frac{dy}{dx}$$-1$
$\Rightarrow \large\frac{x}{y}\frac{dy}{dx}$$+\log y=\large\frac{dy}{dx}$$-1$
$\rightarrow \large\frac{dy}{dx}\big(\large\frac{x}{y}$$-1)=-(\log y+1)$
$\Rightarrow \large\frac{dy}{dx}$$=\large\frac{-(\log y+1)y}{(x-y)}$
$\Rightarrow \large\frac{(1+\log y)y}{(y-x)}$
Step 2:
From equ(1) we get,
$y-x=x\log y$
$\Rightarrow x(1+\log y)=y$
Substituting for $y$ we get,
$\large\frac{dy}{dx}=\frac{x(1+\log y)^2}{x(1+\log y-x)}$
$\Rightarrow \large\frac{x(1+\log y)^2}{x\log y}$
$\Rightarrow \large\frac{(1+\log y)^2}{\log y}$
Hence proved.
answered Jul 2, 2013 by sreemathi.v
 

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