Question

If $x^y+y^x=a^b$ then find that $\frac{dy}{dx}$

• A. $-\frac{y{x}^{x-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}$
• B. $\frac{y{x}^{x-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}$
• C. $-\frac{y{x}^{x-1}+y^x\log y}{x^y\log x+x{y}^x}$
• D. None of these

Answer 1

The correct option is A $-\frac{y{x}^{x-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}$

Let $u=x^y$ and $v=y^x$

$u=x^y$

$\log x=y\log x$

Differentiating we get

$\frac{1}{u}.\frac{du}{dy}=\frac{y}{x}+\log \frac{dy}{dx}$

$\frac{du}{dx}=u(\frac{y}{x}+\log x.\frac{dy}{dx})$

$v=y^x$

$\log v=x\log y$

Differentiating we get

$\frac{1}{v}.\frac{dv}{dx}=\log y+\frac{x}{y}.\frac{dy}{dx}$

$\frac{dv}{dx}=v(\log y+\frac{x}{y}.\frac{dy}{dx})$

$u+v=a^2$

Differentiating, we egt

$\frac{du}{dx}+\frac{dv}{dx}=0$

$\frac{x^y.y}{x}+x^y\log x.\frac{dy}{dx}+y^x\log y+y^x\frac{x}{y}\frac{dy}{dx}=0$

$\frac{dy}{dx}(x^y\log x+x{y}^{x-1})=-(y{x}^{y-1}+y^x\log y)$

$\frac{dy}{dx}=-\frac{y{x}^{y-1}+y^x\log y}{x^y\log x+x{y}^{x-1}}$