Question

If $x^y+y^x=1$ then $\left(\frac{dy}{dx}\right)$ equals

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

Answer 1

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

$x^y+y^x=1\Rightarrow e^{y\log x}+e^{x\log y}=1$

Now differentiating w.r.t $x$

$\Rightarrow e^{y\log x}(\frac{y}{x}+\log x\frac{dy}{dx})+e^{x\log y}(\log y+\frac{x}{y}\frac{dy}{dx})=0$

$\Rightarrow x^y(\frac{y}{x}+\log x\frac{dy}{dx})+y^x(\log y+\frac{x}{y}\frac{dy}{dx})=0$

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