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

1 Answer

  • $(uv)'=u'v+uv'$
Step 1:
Given : $ y^{\large x} = x^{\large y} $
Take $\log$ on both sides
$\log y^{\large x} = \log x^{\large y} $
$\Rightarrow x\log y=y\log x$
Step 2:
Differentiating with respect to $x$
$1.\log y+x.\large\frac{1}{y}\frac{dy}{dx}=\big(\frac{dy}{dx}\big)$$\log x+y.\large\frac{1}{x}$
$\log y+\large\frac{x}{y}\frac{dy}{dx}=$$\log x\large\frac{dy}{dx}+\large\frac{y}{x}$
$\large\frac{x}{y}\frac{dy}{dx}-$$\log x\large\frac{dy}{dx}=\large\frac{y}{x}-$$\log y$
$\Rightarrow \big(\large\frac{x}{y}-$$\log x\big)\large\frac{dy}{dx}=\large\frac{y}{x}$$-\log y$
$\Rightarrow \big(\large\frac{x-y\log x}{y}\big)\frac{dy}{dx}=\large\frac{y-x\log y}{x}$
$\large\frac{dy}{dx}=\large\frac{y(y-x\log y)}{x(x-y\log x)}$
answered May 9, 2013 by sreemathi.v