Inverse Functions

You can use implicit differentiation to find the derivative of inverse functions.

An inverse function reverses the action of another function such that

\[ x = f^{-1}(f(x)) \]

So plugging the result of \( f(x) \) into \( f^{-1}(x) \) will return the original value of \( x \).

You can express this relationship in terms of \( x \) and \( y \) as

\[ f(x) = y \]

\[ g(y) = x \]

So \( f \) and \( g \) are functions that are the inverse of one another.

If you were to graph \( f(x) \) and \( g(y) \) on the same graph it would just be the same line because the \( f \) turns \( x \) values in \( y \) values and \( g \) function does the opposite. Instead we want to exchange the role of \( x \) and \( y \) for the purposes of visualisation by plugging the \( x \) value into the \( g \) function.

For example, if have the function \( f(x) = \sqrt{x} \) and the inverse function \( g(y) = y^2 \) then plugging in \( x \) to \( g \) is the same as the reflecting the curse of \( \sqrt{x} \) across the line \(y = x \). Indeed any inverse function obeys this reversal across this axis.

This exchange of \( x \) and \( y \) is like flipping a fraction where \( \frac{x}{y} \) becomes \( \frac{1}{\frac{x}{y}} \) which is \( \frac{y}{x} \). With simple straight lines like \( y = 2x \) it is plain to see that the inverse of this is \( x = \frac{y}{2} \) so the gradients of these two lines are \( 2 \) and \( \frac{1}{2} \) respectively which is just the reciprocal.

If we consider the definition of the derivative is

\[ \frac{dx}{dy} \]

Then it seems based on the mirroring of inverse graphs that the inverse derivative should be

[\ \frac{1}{\frac{dx}{dy}} \]

\[ \frac{dy}{dx} \]

Proof

We can prove this by using implicit differentiation and the chain rule as well as nesting the inverse function inside the original.

First define the functions.

\[ y = f(x) \]

\[ f^{-1}(y) = x \]

Next take the derivative in terms of \( x \) of \( f^{-1}(y) \)

\[ \frac{d}{dx}(f^{-1}(y)) \]

This is the same as

\[ \frac{d}{dx}(x) \]

Which is equal to \( 1 \) because \( x \) is just a position on the graph which is a constant and the derivative of a constant is always \( 1 \). So

\[ \frac{d}{dx}(f^{-1}(y)) = \frac{d}{dx}(x) = 1 \]

If we the take derivative now using the chain rule, we get the derivative of the outer function \( (f^{-1}(y)) \) as a derivative in terms of the inner function \( y \) or \( \frac{d}{dy} \) the derivative in terms of \( y \). Which is multiplied by derivative of \( y \) in terms of \( x \).

\[ \frac{d}{dy}(f^{-1}(y)) \cdot \frac{d}{dx}y = 1 \]

We can then divide by \( \frac{d}{dx}y \) to show that

\[ \frac{d}{dy}(f^{-1}(y)) = \frac{d}{dx}y = \frac{1}{\frac{d}{dx}y} \]

Which proves the original conjecture of this proof that derivative of the inverse function in terms of \( y \) is simply the derivative of the original function flipped.