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$\frac{d}{dx}e^x = e^x$ 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 $\ln x$.

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Is this a homework question? What have you tried? – Bey Apr 10 '14 at 2:16
It's an extra credit question. I've tried using the composition of functions. – user142118 Apr 10 '14 at 2:17
Okay. What functions did you compose? Do you know how to take the derivative of a composite function? – Bey Apr 10 '14 at 2:19
F(x)=e^x and f^-1(x)=lnx and I think so. I'm kinda just stuck here...I understand d/dx ln x=1/x I'm just can't get the middle steps – user142118 Apr 10 '14 at 2:25
To simplify the notation a bit: let $f(x) = e^x$ and $g(x)=\ln x$. Form the composite function $(f\circ g)(x)=f(g(x))=e^{\ln x}$. Notice that $f(g(x))=x$. Take derivatives on both sides and see what you can do. – Bey Apr 10 '14 at 2:27

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!

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Poster might be interested to know that a derivation for your useful formula exists at:… It is basically an application of the chain rule in the process of taking the derivatives of both sides of the relationship $f(g(x)) = x$, which is of course true if $g$ is the inverse of $f$ – msouth Apr 10 '14 at 5:06

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 $y'=\dfrac{1}{f(f^{-1}(x))} = \boxed{\dfrac1x}$.

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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.

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Don't worry; it was an honest mistake and as such should not have adverse effect on your respected status as burninator. – msouth Apr 10 '14 at 5:01

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 $e^h-1=l$ $$1=\lim_{h\to 0}\frac{e^{h}-1}{h}=\lim_{l\to0}\frac{l}{\ln(l+1)}=\lim_{l\to0}\frac{1}{\ln\sqrt[l]{l+1}}=\frac{1}{\lim_{\,l\to0}\ln\sqrt[l]{l+1}}$$ $$\implies 1=\lim_{l\to0}\ln\sqrt[l]{l+1}\implies \ln^{\prime}(x)=\frac1x$$

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