How do you differentiate $\ln(\exp(z))$

Question

I thought it can be simplified to $\ln(e^z)$

I then thought $e^x$ differentiated is $e^x$ so I thought the $\exp$ part would stay the same

I then thought $\ln x = \frac 1 x$

So I put it all together and decided $\ln(\exp(z))$ derived is $\frac 1 {e^z}$ but I feel this is so wrong.

Answer 1

$$\begin{array}{rcl} \displaystyle \frac{\mathrm d}{\mathrm dz} \ln(\exp(z)) &=& \displaystyle \frac{\mathrm d}{\mathrm d\exp(z)} \ln(\exp(z)) \cdot \frac{\mathrm d\exp(z)}{\mathrm dz} \\ &=& \displaystyle \frac{1}{\exp(z)} \cdot \exp(z) \\ &=& \displaystyle 1 \\ \end{array}$$

On a trivial remark, $\ln(\exp(z)) \equiv z$, so $\displaystyle \frac{\mathrm d}{\mathrm dz} \ln(\exp(z)) = \displaystyle \frac{\mathrm d}{\mathrm dz} z = 1$.