derivative of this special function

Question

I would like to take the first derivative of the following function respect to x. what is the derivative of this function with respect to x?

$$f = {(e^{y-z})}^{e^{xw}}$$ where y, z, and w are known.

Answer 1

One approach is the following: $$ \begin{align} f &= (e^{y-z})^{e^{wx}}\\ &= e^{(y-z)e^{wx}}\\ \frac{\partial f}{\partial x} &= e^{(y-z)e^{wx}} \frac{d}{dx}[(y-z)e^{wx}] \quad\text{by the chain rule}\\ &= e^{(y-z)e^{wx}} [(y-z)we^{wx}]\quad\text{because $(y-z)$ is a constant}\\ &= w(y-z)e^{wx}e^{(y-z)e^{wx}} \end{align} $$

Answer 2

Another way would be to write, from $$f = {(e^{y-z})}^{e^{xw}}$$ $$\log(f) = e^{xw} (y-z)$$ and then, by differentiation of both sides, $$\frac{f_x'}{f}=we^{xw} (y-z)$$ and then $f_x'$