If $a^x+a^y=a^{x+y}$, then $\frac{dy}{dx}=$

$a^x+a^y=a^{x+y}$

Differentiate  with respect to $\text{'}x\text{'}$

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

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

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

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

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