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Find $\large\frac{dy}{dx}$ of the functions given in $ x^{\large y }+ y^{\large x }= 1 $

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Toolbox:
  • $\log m^{\large n}=n\log m$
Step 1:
Let $u=x^{\large y}$, $v=y^{\large x}$
$u+v=1$
$\large\frac{du}{dx}+\frac{dv}{dx}$$=0$-------(1)
Now $u=x^{\large y}$
Taking $\log$ on both sides
Now $\log u=\log x^{\large y}$
$\log m^{\large n}=n\log m$
$\Rightarrow \log u=y\log x$
Step 2:
Differentiating with respect to $x$
$\large\frac{1}{u}\frac{du}{dx}=\big(\large\frac{dy}{dx}\big)$$\log x+y.\large\frac{1}{x}$
$(uv)'=u'v+uv'$
$\large\frac{du}{dx}=$$u[\big(\large\frac{dy}{dx}\big)$$\log x+\large\frac{y}{x}]$
$\quad=x^{\large y}[(\log x)\large\frac{dy}{dx}+\frac{y}{x}]$
Step 3:
Consider $v=y^{\large x}$
Taking $\log$ on both sides
$\log v=\log y^{\large x}$
$\log m^{\large n}=n\log m$
$\log v=x\log y$
Differentiating with respect to $x$
$\large\frac{1}{v}\frac{dv}{dx}$$=1.\log y+x.\large\frac{1}{y}\frac{dy}{dx}$
$\large\frac{dv}{dx}$$=v(\log y+\large\frac{x}{y}\frac{dy}{dx})$
Step 4:
Substitute the value of $\large\frac{du}{dx}$ and $\large\frac{dv}{dx}$ in eq(1)
$x^{\large y}[(\log x)\large\frac{dy}{dx}+\frac{y}{x}]$$+y^{\large x}[\log y+\large\frac{x}{y}\frac{dy}{dx}]$$=0$
$\Rightarrow (x^y\log x+xy^{\large{x-1}})\large\frac{dy}{dx}$$+yx^{\large{y-1}}+y^{\large x}\log y$=0
$\Rightarrow (x^y\log x+xy^{\large{x-1}})\large\frac{dy}{dx}$$=-yx^{\large{y-1}}+y^{\large x}\log y$
$\Large\frac{dy}{dx}=\Large\frac{-yx^{\Large {y-1}}+y^{\Large x}\log y}{x^{\Large y}\log x+xy^{\Large{x-1}}}$
answered May 9, 2013 by sreemathi.v
 

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