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

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:
1. the conditions under which the inverse function, $$g$$, is differentiable;
2. the formula of $$g’$$.

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

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:

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

Example 1

Let $$f(x)=2x^{3}+3x^{2}+6x+1$$. Find $$(f^{-1})'(1)$$ if it exists.

Solution

The function $$f$$ is a polynomial and hence differentiable everywhere. Also because
$f'(x)=6x^{2}+6x+6=6(x^{2}+x+1)=6\left[\left(x+\frac{1}{2}\right)^{2}+\frac{3}{4}\right]>0,$
its inverse is differentiable everywhere too. It follows from the Derivative Rule for Inverses that
$(f^{-1})'(1)=\frac{1}{f'(f^{-1}(1))}.$ To find $$f^{-1}(1)$$, we need to solve $$f(x)=1$$ or $2x^{3}+3x^{2}+6x+1=1.$ We can see that $$x=0$$ is a solution to this equation. Because $$f$$ has to be one-to-one to have an inverse, $$x=0$$ must be the only solution. So $(f^{-1})'(1)=\frac{1}{f'(0)}$ and to calculate $$f'(0)$$, we just plug $$x=0$$ into $$f'(x)$$ we already obtained. That is, $f'(0)=\left.6x^{2}+6x+6\right|_{x=0}=6$ and finally

$(f^{-1})'(1)=\frac{1}{f'(0)}=\frac{1}{6}.$

Example 2

Let $$y=f(x)=x^{2}+2x$$ for $$x>-1$$. Find $$\dfrac{d}{dx}f^{-1}(x)$$.

Solution

Method 1: Let’s find $$f^{-1}$$ and then differentiate it. To find $$f^{-1}$$, we start off with the equation $y=x^{2}+2x$ and solve it for $$x$$: $x=\frac{-2\pm\sqrt{4-4y}}{2}=-1\pm\sqrt{1+y}.$ This equation gives two values of $$x$$ for each $$y$$, but we have to choose the one with the $$+$$ sign because it is assumed that $$x>-1$$. Therefore $x=f^{-1}(y)=-1+\sqrt{1+y}.$ In the equation $$f^{-1}(y)=-1+\sqrt{1+y}$$, $$y$$ is a variable that shows the input. We can denote the input with whatever we want, including $$x$$. So $f^{-1}(x)=-1+\sqrt{1+x}.$ Now we can easily find $$(f^{-1})'(x)$$:
$\frac{d}{dx}f^{-1}(x)=\frac{1}{2\sqrt{1+x}}.$
Method 2: Start off with $$y=f(x)$$: $\frac{df}{dx}=\frac{dy}{dx}=2x+2.$ By the Derivative Rule for inverses, we have
\begin{align} \frac{df^{-1}}{dy}=\frac{dx}{dy} & =\frac{1}{dy/dx}\\ & =\frac{1}{2x+2}\end{align}
Now we need to express $$dx/dy$$ in terms of $$y$$ because in $$df^{-1}/dy=f'(y)$$, the independent variable is $$y$$. From $$y=x^{2}+2x$$, we solve for $$x$$:
\begin{align} x & =\frac{-2\pm\sqrt{4+4y}}{2}\\ & =-1\pm\sqrt{1+y}.\end{align}
Because $$x>-1$$, we set $$x=-1+\sqrt{1+y}$$. Therefore
\begin{align} \frac{df^{-1}}{dy} & =\frac{1}{2(-1+\sqrt{1+y})+2}\\ & =\frac{1}{2\sqrt{1+y}}.\end{align}
Here $$y$$ simply shows the input of $f^{-1}$; we can replace it by $$x$$ and write:
$\frac{df^{-1}}{dx}=\left(f^{-1}\right)'(x)=\frac{1}{2\sqrt{1+x}}.$

Example 3

Suppose $$h(x)=\tan(f^{-1}(x))$$ and we know $$f\left(\dfrac{\pi}{4}\right)=1$$ and $$f’\left(\dfrac{\pi}{4}\right)=5$$. Find $$h'(1)$$.

Solution

Let $$u=f^{-1}(x)$$. Thus $$h(x)=\tan u$$ and using the chain rule, we get
\begin{align} h'(x) & =u’\left(\frac{d}{du}\tan u\right)\\ & =u’\cdot(1+\tan^{2}u)\\ & =(f^{-1})'(x)\cdot\left[1+\tan^{2}(f^{-1}(x))\right].\end{align}
Because the Derivative Rule for Inverses tells us
$(f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))},$ we have $h'(x)=\frac{1}{f'(f^{-1}(x))}\left[1+\tan^{2}(f^{-1}(x))\right]$ and
$h’\left(1\right)=\frac{1}{f'(f^{-1}(1))}\left[1+\tan^{2}(f^{-1}(1))\right].$
Because $$f(\pi/4)=1$$ means $$f^{-1}(1)=\pi/4$$, we obtain
\begin{align} h'(1) & =\frac{1}{f'(\pi/4)}\left[1+\tan^{2}(\pi/4)\right]\\ & =\frac{1}{5}\left[1+1^{2}\right]\\ & =\frac{2}{5}.\end{align}