Reciprocal relation between derivatives of Exp(x) and Log(x)

Question

We know that $\exp x=e^x$ and $\ln x$ are inverse functions of each other, i.e. if $y=\exp x$ then $x=\ln y$. Let $f(x)=\exp x$ and $g(y)=\ln y$, then the derivatives of these functions are $f'(x)=e^x$ and $g'(y)=\frac 1y$ respectively.

Now, a theorem of Calculus states that when $f(x)$ and $g(y)$ are inverse functions then $g'(y)=\frac 1{f'(x)}$. This theorem is used to determine the derivatives of the inverse trig functions.

Can someone indicate how this theorem relates $f'(x)=e^x$ and $g'(y)=\frac 1y$? There should be an underlying reciprocal relation between the two functions which I fail to see.

Answer 1

we have $$\frac{dy}{dx}\cdot \frac{dx}{dy}=1$$ for example $$y=e^x$$ we get the inverse function as $$x=\ln(y)$$ and the derivative is $$\frac{dx}{dy}=\frac{1}{y}=\frac{1}{e^x}$$ thus we get $$\frac{dy}{dx}\cdot \frac{dx}{dy}=e^x\cdot \frac{1}{e^x}=1$$

Answer 2

The Inverse Function Theorem states for a function $f$ and its inverse $f^{-1}$ at a point $y=f(a)$:

$$\left(f^{-1}\right)'(f(a))=\frac{1}{f'(a)}$$

If you set $f(x)=\exp(x)$ and $g(y)=f^{-1}(y)=\ln(y)$, then

$$g'(f(a))=\frac{1}{f'(a)}\Leftrightarrow$$ $$\ln'(f(a))=\frac{1}{\exp'(a)}\Leftrightarrow$$ $$\frac{1}{f(a)}=\frac{1}{\exp(a)}\Leftrightarrow$$ $$\exp(a)=\exp(a)$$

which is true.

Answer 3

The graph of a function and its inverse are equally inclined at the same angle say $\phi= \tan ^{-1}t $ symmetrical (mirrored) about line $ x=y.$

$ \tan ( \phi + \pi/4) = \dfrac{1+t}{1-t} $ for one graph and $ \tan ( \phi - \pi/4) = \dfrac{1-t}{1+t} $ for its inverse.

Although you know about the basic reciprocal relation of slopes, know its proof but still unable to appreciate the same for some pairs, I am posting graphs of the two inversive pairs $ ( e^x,\log x), (x^2, \sqrt x). $ This can enable computing some points along with their slopes to check the theoretical result to be correct, across the inverting points either way. They enter into a inverting relation seen as a geometric reflection.

Points (0,1) of $ e^x$ goes to (1,0) of $ \log x.$ The slopes are same,$=1.$ A short tangent line is marked at all these points, (1&2).

Another point pair (5&6) chosen is $(e,1)$ and its mirror $ (1,e) $ chosen on same log and exponential curves.

For parabolas $ y=x^2, y=\sqrt x $ the intersection point is $(1,1)$ with slopes $(2,1/2).$ for points (3&4).

Answer 4

A result of the definition of an inverse function of a 1-1 function is:

$$f(f^{-1}(x))=x$$

Applying implicit differentiation (and keeping in mind the chain rule) with respect to $x$ on both sides we get:

$$f'(f^{-1}(x))•(f^{-1}(x))'=1$$

This is basically the theorem your given but I think it's more clear.

The derivative of $e^x=e^x$

Thus:

$$f'(f^{-1}(x))=e^{lnx}=x$$

And the derivative of $lnx=\frac{1}{x}$

If you multiply these you will get one, as showed by our equation.

The two slopes of the tangents in the picture should multiply to 1. The green line is $y=x$ and the purple is a line perpendicular to $y=x$.