# Derivative of $e^x$ after geometric transformation

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?

• 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? – Dunnò000 Jun 13 at 13:26

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 $$e^{\ln(x)}=x$$: $$x\cdot \frac{d}{dx}\ln(x) = 1$$ $$\frac{d}{dx}\ln(x) = \frac{1}{x}$$
• 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? – 1b3b Jun 13 at 12:05
• The slope is $1/e^{\ln(a)}=1/a$ – Integrand Jun 13 at 12:06