Proof of the derivative of ln(x)

If you can use the chain rule and the fact that the derivative of $e^x$ is $e^x$ and the fact that $\ln(x)$ is differentiable, then we have:

$$\frac{\mathrm{d} }{\mathrm{d} x} x = 1$$

$$\frac{\mathrm{d} }{\mathrm{d} x} e^{\ln(x)} = e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$e^{\ln(x)} \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$x \frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = 1$$

$$\frac{\mathrm{d} }{\mathrm{d} x} \ln(x) = \frac{1}{x}$$

The simpler way is to use the inverse function theorem for derivatives:

If $f$ is a bijection from an interval $I$ onto an interval $J=f(I)$, which has a derivative at $x\in I$, and if $f'(x)\neq 0$, then $f^{-1}\colon J\to I$ has a derivative at $y=f(x)$, and $$\bigl(f^{-1}\bigr)'(y)=\frac1{f'(x)}=\frac1{f'\bigl(f^{-1}(y)\bigr)}.$$

As $(\mathrm e^x)'=\mathrm e^x\neq 0\,$ for all $x$, we know that $\,\ln\,$ has a derivative at each point of its domain, and $$(\ln)'(y)=\frac1{\mathrm e^{\,\ln y}}=\frac1y.$$

Define $$e=\lim_{h\to 0} \left(1+h\right)^{1/h}.$$ Then change variables $h\mapsto h/x$ giving $$e=\lim_{h/x\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}}=\lim_{h\to 0} \left(1+\frac{h}{x}\right)^{\frac{x}{h}},$$ where the limit in the second equality follows since $h$ approaches $0$ as $h/x$ does. Since $x$ is constant w.r.t. $h$, we can simplify by raising both sides to the power $1/x$, giving you the desired identity.