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Inverse function of $f(x) = e^x$ is of course $f^{-1}(x) = \ln{x}.$ We have, by definition, $\frac{d}{dx}e^x = e^x$. In other words, $e^x$ in some sense describes slopes of tangent lines on a curve given by outputs of $e^x$.

We can get $\ln{x}$ by reflecting curve $e^x$ over the line $y = x$. But this is equvalent to rotate Cartesian plane by $90°$ and then reflect by "new" vertical axis ($x$ axis). By this transformation is easy to see that slope $m$ of any line in $xy$ plane is now, after this transformation, equal $\frac{1}{m}$ in $yx$ plane.

Now, I concluded: Since slopes are reciprocal it must be $\frac{d}{dx} \ln{x} = \frac{1}{e^x}$. But I know that $\frac{d}{dx} \ln{x} = \frac{1}{x}.$ Obviously $\frac{1}{x} \neq e^x.$ Where am I mistaking?

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  • $\begingroup$ By your reasoning, the slope of the tangent at the point $(x, \ln(x))$ of $\{(x,y): y=\ln(x)\}$ equals the inverse of the slope of the tangent at the poin $(\ln(x), x)$ of $\{(x,y): y=e^x\}$. Can you see your mistake from this? $\endgroup$ – Dunnò000 Jun 13 at 13:26
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A heuristic explanation, omitting a few details.

Claim: $e^x$ is differentiable, increasing, and has range $(0,\infty)$; therefore $\ln(x)$ is differentiable on $(0,\infty)$. Further, as you pointed out, $e^x$ and $\ln(x)$ are function inverses: on their natural domains, $e^{\ln(x)}=x$ and $\ln(e^x)=x$.

Say a priori we had no idea what $\frac{d}{dx}\ln(x)$ was, but we knew about the Chain Rule. Then $$ e^{\ln(x)}=x $$ $$ \frac{d}{dx}e^{\ln(x)}=\frac{d}{dx}x $$ $$ e^{\ln(x)}\cdot \frac{d}{dx}\ln(x)= 1 $$Now use the fact that $e^{\ln(x)}=x$: $$ x\cdot \frac{d}{dx}\ln(x) = 1 $$ $$ \frac{d}{dx}\ln(x) = \frac{1}{x} $$

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    $\begingroup$ This is not bad answer, but my question was where I made mistake in reasoning? Consider some $x = a.$ Slope of tangent line on $e^x$ in point $(a, e^x)$ is $e^a.$ By analysis given above, slope on point $(a, \ln{x})$ is $\frac{1}{e^a}$. Where is flawed logic? $\endgroup$ – 1b3b Jun 13 at 12:05
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    $\begingroup$ The slope is $1/e^{\ln(a)}=1/a$ $\endgroup$ – Integrand Jun 13 at 12:06
  • $\begingroup$ Oh yes, derivative changes after first transformation and after second. $\endgroup$ – 1b3b Jun 13 at 12:10

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