Question

If $x^y+y^x=a^b$ the find $\frac{dy}{dx}$. OR
If $e^y(x+1)=1,$ then show that $\frac{d^{2}y}{d{x}^2}={\left(\frac{dy}{dx}\right)}^2$.

Answer 1

We have $x^y+y^x=a^b\Rightarrow e^{log{x}^y}+e^{log{y}^x}=a^b\Rightarrow e^{ylogx}+e^{xlogy}=a^b$

$\Rightarrow e^{ylogx}(y\times \frac{1}{x}+logx\times \frac{dy}{dx})+e^{xlogy}(x\times \frac{1}{y}\times \frac{dy}{dx}+logy.1)=0$

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

$\Rightarrow x^{y-1}y+x^{y}logx\times \frac{dy}{dx}+y^{x-1}x\times \frac{dy}{dx}+y^{x}logy=0$

$\Rightarrow (x^{y}logx+y^{x-1}x)\frac{dy}{dx}=-(y^{x}logy+x^{y-1}y)\therefore \frac{dy}{dx}=-\left(\frac{y^{x}logy+x^{y-1}y}{x^{y}logx+y^{x-1}x}\right)$

OR We have $e^y(x+1)=1\Rightarrow e^y=\frac{1}{x+1}\Rightarrow lo{g}_e{e}^y=lo{g}_e\left(\frac{1}{x+1}\right)or,y=-log(x+1)$

$Now\frac{dy}{dx}=-\frac{1}{x+1}...\left(i\right)and,\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(-\frac{1}{x+1})$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{1}{(x+1{)}^2}={(-\frac{1}{x+1})}^2={\left(\frac{dy}{dx}\right)}^2\left[By\right(i)$

Therefore,$\frac{d^{2}y}{d{x}^2}={\left(\frac{dy}{dx}\right)}^2$.