Question

Differentiate w.r.t $x$ :$x^y+y^x=1$

Answer 1

Let $x^y=t$ ......$\left(1\right)$

$\Rightarrow y\log x=\log t$ by taking $\log$ both sides.

Differentiating w.r.t $x$ we get

$\Rightarrow \frac{y}{x}+\log x\frac{dy}{dx}=\frac{1}{t}\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}=t(\frac{y}{x}+\log x\frac{dy}{dx})$

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

$\Rightarrow \frac{dt}{dx}=y{x}^{y-1}+x^y\log x\frac{dy}{dx}$ ......$\left(2\right)$

Let $y^x=w$ .....$\left(3\right)$

$\Rightarrow x\log y=\log w$ by taking $\log$ both sides.

Differentiating w.r.t $x$ we get

$\Rightarrow \frac{x}{y}\frac{dy}{dx}+\log y=\frac{1}{w}\frac{dw}{dx}$

$\Rightarrow \frac{dw}{dx}=w(\frac{x}{y}\frac{dy}{dx}+\log y)$

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

$\Rightarrow \frac{dw}{dx}=x{y}^{x-1}\frac{dy}{dx}+y^x\log y$ ......$\left(4\right)$

From the question we have $x^y+y^x=1$

$\Rightarrow t+w=1$ from $\left(1\right)$ and $\left(3\right)$

$\Rightarrow \frac{dt}{dx}+\frac{dw}{dx}=0$

$\Rightarrow y{x}^{y-1}+x^y\log x\frac{dy}{dx}+x{y}^{x-1}\frac{dy}{dx}+y^x\log y=0$ from $\left(2\right)$ and $\left(4\right)$

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

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