*In this section, we investigate the relationship between the derivative of a function and the derivative of its inverse function.*

Let \(f\) be a one-to-one and differentiable function. Because it is differentiable, its graph does not have any corners or any cusps. Because the graph of \(f^{-1}\) is obtained by reflecting the graph of $f$ across the line \(y=x\), the graph of \(f^{-1}\) does not have any corners or any cusps either. We, therefore, expect that \(f^{-1}\) is differentiable wherever the tangent to its graph is not vertical.

Consider a point \(\left(a,f(a)\right)\) on the graph of \(f\) (Figure 1). The equation of the tangent line \(L\) to the graph of \(f\) at \((a,f(a))\) is

\[L:\quad y-f(a)=f'(a)(x-a).\]

If the graph of \(f\) and the line \(L\) are reflected through the diagonal line \(y=x\),the graph of \(f^{-1}\) and the tangent line \(L’\) through \((f(a),a)\) are obtained (Figure 2). The equation of \(L’\) is obtained from the equation of \(L\) by interchanging \(x\) and \(y\); that is, \[L’:\quad x-f(a)=f'(a)(y-a)\] or

\[L’:\quad y-a=\frac{1}{f'(a)}(x-f(a)),\]
which shows that the slope of $L’$ is $1/f'(a)$. On the other hand, because $L’$ is the tangent line to the graph of $f^{-1}$ at $\left(f(a),a\right)$, the slope of $L’$ is $\left(f^{-1}\right)’\left(f(a)\right)$. Therefore,

\[\left(f^{-1}\right)’\left(f(a)\right)=\frac{1}{f'(a)}.\]
Let $b=f(a)$. Then because $a=f^{-1}(b)$, the above equation can be alternatively written as

\[\left(f^{-1}\right)^{\prime}(\underbrace{b}_{f(a)})=\frac{1}{f'(\underbrace{f^{-1}(b)}_{a})}.\]

In general,

\[\bbox[#F2F2F2,5px,border:2px solid black]{(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}}\]

for every \(y\) in the domain of \(f^{-1}\).

**Theorem 1. (The Derivative Rule for Inverses):** Assume that \(f\) is a function which is differentiable on the interval \((a,b)\) such that \(f'(x)>0\) (or \(f'(x)<0\)) for all \(x\) in \((a,b).\) Then the inverse function \(x=g(y)\) exists, and we have

\[g'(y)=\frac{1}{f'(x)}=\frac{1}{f'(g(y))},\] or \[g’=\frac{1}{f’\circ g}.\]

- In the Leibniz notation, \[x=g(y)\Rightarrow g'(y)=\frac{dx}{dy},\] and \[y=f(x)\Rightarrow f'(x)=\frac{dy}{dx}.\] So the above theorem when expressed in the Leibniz notation becomes

\[\frac{dx}{dy}=\frac{1}{\dfrac{dy}{dx}}.\] Notice that in $dx/dy$ , the independent variable is $y$ , and in $dy/dx$ , the independent variable is $x$ . Additionally notice that if $(x_0 , y_0 )$ is on the graph of $f$ , then $dy/dx$ must be evaluated at $x_0$ and $dx/dy$ must be evaluated at $y_0$, or\[\bbox[#F2F2F2,5px,border:2px solid black]{\left(\frac{dx}{dy}\right)_{y_{0}}=\frac{1}{\left(\dfrac{dy}{dx}\right)_{x_{0}}}.}\]

#### Show the proof

#### Hide the proof

Roughly speaking, because

\[\frac{\Delta x}{\Delta y}=\frac{1}{\dfrac{\Delta y}{\Delta x}}\] and \(\Delta y\to0\) as \(\Delta x\to0\), we have

\[g'(y)=\lim_{\Delta y\to0}\frac{\Delta x}{\Delta y}=\frac{1}{\underset{\Delta x\to0}{\lim}\dfrac{\Delta y}{\Delta x}}=\frac{1}{f'(x)}.\]

However, for a rigorous mathematical proof we need to include more steps and details.

- The above theorem makes two assertions:

- the conditions under which the inverse function, \(g\), is differentiable;
- the formula of \(g’\).

#### Show an alternative proof for the above theorem

#### Hide the alternative proof

If it is known that \(g\) is differentiable, we can derive the formula of \(g’\) using implicit differentiation in the following way. Let’s start off with \[x=g(y)\] Then implicitly differentiate \(g(y)=x\) with respect to \(x\): \[\frac{d}{dx}(g(y))=\frac{d}{dx}(x)\] The right hand side is one, and for the left hand side we use the chain rule:

\[\underbrace{\frac{dg}{dy}}_{g'(y)}\underbrace{\frac{dy}{dx}}_{f'(x)}=1.\]

Therefore \[g'(y)=\frac{dx}{dy}=\frac{1}{f'(x)}=\frac{1}{dy/dx}.\]

To show how the above formula works, consider \(y=f(x)=x^{2}\). The inverse function is \(x=f^{-1}(y)=\sqrt{y}\). Because \[\frac{dy}{dx}=\frac{df}{dx}=2x,\] we have

\[\begin{align} \frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x}\\ & =\frac{1}{2\sqrt{y}}.\end{align}\]

Thus \((f^{-1})'(y)=\dfrac{1}{2\sqrt{y}}\). Here $y$ is the independent variable*, *but if we wish, we can denote the independent variable, as usual, by $x$. To do so, we can simply replace $y$ with $x$ on both sides of the equation:

\[(f^{-1})'(x)=\frac{1}{2\sqrt{x}}.\]