Question

Given that ${\left(f\left(x\right)\right)}^{g\left(y\right)}=e^{f\left(x\right)-g\left(y\right)}$. Find $\frac{df\left(x\right)}{dg\left(x\right)}$

Answer 1

$\left(f\right(x){)}^{g\left(x\right)}=e^{f\left(x\right)-g\left(x\right)}$

Apply logarithm on both sides

$g\left(x\right)\ln f\left(x\right)=f\left(x\right)-g\left(x\right)$

Differentiating with respect to $g\left(x\right)$

$\ln f\left(x\right)+\frac{g\left(x\right)}{f\left(x\right)}\frac{df\left(x\right)}{dg\left(x\right)}=\frac{df\left(x\right)}{dg\left(x\right)}-1$

$\frac{df\left(x\right)}{dg\left(x\right)}=\frac{\ln f\left(x\right)+1}{1-\frac{g\left(x\right)}{f\left(x\right)}}=\frac{f\left(x\right)\ln f\left(x\right)+f\left(x\right)}{f\left(x\right)-g\left(x\right)}$