Question

If $x^2=c^{x-y},$ then find $\frac{dy}{dx}$

Answer 1

$x^2=c^{x-y}$

$2\log x=(x-y)\log c=x\log c-y\log c$

Differentiating both sides.

$\frac{2}{x}=\log c-\log c\frac{dy}{dx}$

$\log c\frac{dy}{dx}=\log c-\frac{2}{x}=\frac{x\log c-2}{x}$

$\frac{dy}{dx}=\frac{1}{\log c}\left(\frac{x\log x-2}{x}\right)=1-\frac{2}{x\log c}$