Find \frac{dy}{dx},\: if\: x^{y}\cdot y^{x}= x^{x}\cdot

 

 

 

 
 
 
 
 

Answers (1)

x^{y}\cdot y^{x}= x^{x}
Taking log on both sides we get
\log \left ( x^{y} \cdot y^{x}\right )= \log x^{x}
\log x^{y} +\log y^{x}= x\log x
y\log x+x\log y = x\log x  [by the law of Expanant ]
Differentiating the above equation
\left [ \frac{y}{x}+\log x\frac{dy}{dx} \right ]+\left [ \frac{x}{y}\frac{dy}{dx}+\log y \right ]= 1+\log x\:
                                                               \left [ \because {f}' \left ( uv \right )= u{f}'\left ( v \right )+v{f}'\left ( u \right )\right ]
\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y= 1+\log x
\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}= 1+\log x-\log y-\frac{y}{x}
\therefore \frac{dy}{dx}= \frac{x+x\log x-x\log y-y}{x\left ( \log x+\frac{x}{y} \right )}
\therefore \frac{dy}{dx}= \frac{y\left ( x+x\log x-x\log y -y \right )}{\left ( y\log +x \right )x}
 

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