Derive an equation for derivative of ln x

Question

$\frac{d}{dx}e^x = e^x$ use this fact together with the definition of the natural log $\ln x$ as the inverse of the function of $e^x$ to derive an equation for the derivative of $\ln x$.

Answer 1

A useful formula to know is the following: $$f^{\prime}(x) = \frac{1}{(f^{-1})^{\prime}f(x)}$$ Plug in functions ($(f^{-1})^{\prime} = e^x$, $f(x) = \ln(x)$) to get the following: \begin{align*} \ln^{\prime}(x) & = \frac{1}{e^{\ln(x)}} \\ & = \frac{1}{x} \text{.} \end{align*}

Voila, a formula for the derivative of the natural logarithm!

Answer 2

Try setting $y=e^x$, so $\frac{dy}{dx} = y$. Now, solve this ODE via separation of variables.

EDIT: Never mind, did not see the part about using the inverse.

Answer 3

If $y=f^{-1}(x)$ then $f(y)=x$, so $f'(y)y'=1$ (chain rule); thus $y'=\dfrac{1}{f'(y)} =\dfrac{1}{f'(f^{-1}(x))}$.

Using the fact that $f'=f$ for the exponential function, this means $y'=\dfrac{1}{f(f^{-1}(x))} = \boxed{\dfrac1x}$.

Answer 4

Just using the definitions: $$\ln^{\prime}(x)=\lim_{h\to 0} \frac{\ln(x+h)-\ln(x)}h=\lim_{h\to 0} \frac{\ln(1+\frac h x)}h=\lim_{h\to 0}\ln\left(\sqrt[h]{1+\frac h x}\right)$$

Substituting $h=xa$ $$\ln^{\prime}(x)=\lim_{a\to 0}\ln\left(\sqrt[xa]{1+a}\right)=\lim_{a\to 0}\ln\left(\sqrt[a]{1+a}\right)/x$$ Now, using that the derivative of $e^x$ is $e^x$ $$e^x=\lim_{h\to 0}\frac{e^{x+h}-e^x}{h}\implies 1=\lim_{h\to 0}\frac{e^{h}-1}{h}$$ Substituting $e^h-1=l$ $$1=\lim_{h\to 0}\frac{e^{h}-1}{h}=\lim_{l\to0}\frac{l}{\ln(l+1)}=\lim_{l\to0}\frac{1}{\ln\sqrt[l]{l+1}}=\frac{1}{\lim_{\,l\to0}\ln\sqrt[l]{l+1}}$$ $$\implies 1=\lim_{l\to0}\ln\sqrt[l]{l+1}\implies \ln^{\prime}(x)=\frac1x$$