5.12 Derivatives of Inverse Functions

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 $(a, f (a))$ on the graph of $f$ (Figure 1). The equation of the tangent line $L$ to the graph of $f$ at $(a, f (a))$ is
$L : y - f (a) = f^{'} (a) (x - a) .$

Figure 1

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, $L^{'} : x - f (a) = f^{'} (a) (y - a)$ or
$L^{'} : y - a = \frac{1}{f^{'} (a)} (x - f (a)),$ 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 $(f (a), a)$ , the slope of $L^{'}$ is ${(f^{- 1})}^{'} (f (a))$ . Therefore,
${(f^{- 1})}^{'} (f (a)) = \frac{1}{f^{'} (a)} .$ Let $b = f (a)$ . Then because $a = f^{- 1} (b)$ , the above equation can be alternatively written as
${(f^{- 1})}^{'} (\underset{f (a)}{\underset{⏟}{b}}) = \frac{1}{f^{'} (\underset{a}{\underset{⏟}{f^{- 1} (b)}})} .$

In general,
$(f^{- 1})^{'} (y) = \frac{1}{f^{'} (f^{- 1} (y))}$
for every $y$ in the domain of $f^{- 1}$ .

Figure 2

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, $x = g (y) \Rightarrow g^{'} (y) = \frac{d x}{d y},$ and $y = f (x) \Rightarrow f^{'} (x) = \frac{d y}{d x} .$ So the above theorem when expressed in the Leibniz notation becomes
$\frac{d x}{d y} = \frac{1}{\frac{d y}{d x}} .$ Notice that in $d x / d y$ , the independent variable is $y$ , and in $d y / d x$ , the independent variable is $x$ . Additionally notice that if $(x_{0}, y_{0})$ is on the graph of $f$ , then $d y / d x$ must be evaluated at $x_{0}$ and $d x / d y$ must be evaluated at $y_{0}$ , or

${(\frac{d x}{d y})}_{y_{0}} = \frac{1}{{(\frac{d y}{d x})}_{x_{0}}} .$

Show the proof

Roughly speaking, because
$\frac{Δ x}{Δ y} = \frac{1}{\frac{Δ y}{Δ x}}$ and $Δ y \to 0$ as $Δ x \to 0$ , we have
$g^{'} (y) = lim_{Δ y \to 0} \frac{Δ x}{Δ y} = \frac{1}{lim_{Δ x \to 0} \frac{Δ y}{Δ x}} = \frac{1}{f^{'} (x)} .$
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

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 $x$ : $\frac{d}{d x} (g (y)) = \frac{d}{d x} (x)$ The right hand side is one, and for the left hand side we use the chain rule:
$\underset{g^{'} (y)}{\underset{⏟}{\frac{d g}{d y}}} \underset{f^{'} (x)}{\underset{⏟}{\frac{d y}{d x}}} = 1.$
Therefore $g^{'} (y) = \frac{d x}{d y} = \frac{1}{f^{'} (x)} = \frac{1}{d y / d x} .$

To show how the above formula works, consider $y = f (x) = x^{2}$ . The inverse function is $x = f^{- 1} (y) = \sqrt{y}$ . Because $\frac{d y}{d x} = \frac{d f}{d x} = 2 x .$ Then
$\begin{aligned} \frac{d x}{d y} & = \frac{1}{d y / d x} \\ = \frac{1}{2 x} \\ = \frac{1}{2 \sqrt{y}} . \end{aligned}$
Thus $(f^{- 1})^{'} (y) = \frac{1}{2 \sqrt{y}}$ . 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 $f (x) = 2 x^{3} + 3 x^{2} + 6 x + 1$ . Find $(f^{- 1})^{'} (1)$ if it exists.

Solution

The function $f$ is a polynomial and hence differentiable everywhere. Also because
$f^{'} (x) = 6 x^{2} + 6 x + 6 = 6 (x^{2} + x + 1) = 6 [{(x + \frac{1}{2})}^{2} + \frac{3}{4}] > 0,$
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 $2 x^{3} + 3 x^{2} + 6 x + 1 = 1.$ 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, $f^{'} (0) = {6 x^{2} + 6 x + 6 |}_{x = 0} = 6$ and finally

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

Example 2

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

Solution

Method 1: Let’s find $f^{- 1}$ and then differentiate it. To find $f^{- 1}$ , we start off with the equation $y = x^{2} + 2 x$ and solve it for $x$ : $x = \frac{- 2 \pm \sqrt{4^{2} - 4 y}}{2} = - 1 \pm \sqrt{1 + y} .$ 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 $(f^{- 1})^{'} (x)$ :
$\frac{d}{d x} f^{- 1} (x) = \frac{1}{2 \sqrt{1 + x}} .$
Method 2: Start off with $y = f (x)$ : $\frac{d f}{d x} = \frac{d y}{d x} = 2 x + 2.$ By the Derivative Rule for inverses, we have
$\begin{aligned} \frac{d f^{- 1}}{d y} = \frac{d x}{d y} & = \frac{1}{d y / d x} \\ = \frac{1}{2 x + 2} \end{aligned}$
Now we need to express $d x / d y$ in terms of $y$ because in $d f^{- 1} / d y = f^{'} (y)$ , the independent variable is $y$ . From $y = x^{2} + 2 x$ , we solve for $x$ :
$\begin{aligned} x & = \frac{- 2 \pm \sqrt{4 + 4 y}}{2} \\ = - 1 \pm \sqrt{1 + y} . \end{aligned}$
Because $x > - 1$ , we set $x = - 1 + \sqrt{1 + y}$ . Therefore
$\begin{aligned} \frac{d f^{- 1}}{d y} & = \frac{1}{2 (- 1 + \sqrt{1 + y}) + 2} \\ = \frac{1}{2 \sqrt{1 + y}} . \end{aligned}$
Here $y$ simply shows the input of $f^{- 1}$ ; we can replace it by $x$ and write:
$\frac{d f^{- 1}}{d x} = {(f^{- 1})}^{'} (x) = \frac{1}{2 \sqrt{1 + x}} .$

Example 3

Suppose $h (x) = \tan (f^{- 1} (x))$ and we know $f (\frac{π}{4}) = 1$ and $f^{'} (\frac{π}{4}) = 5$ . Find $h^{'} (1)$ .

Solution

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

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Show an alternative proof for the above theorem

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