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E.2 Derivative of Inverse Functions

Let’s now develop a formula that relates the derivative of a function’s inverse to the derivative of the original function. We’ll first use a geometric approach, which offers tremendous insight into why the two derivatives are related as they are. We’ll then use the Chain Rule to develop the same result, and show you a procedure that will always work even if you don’t remember the formula.

Geometric Development of the Inverse Function Theorem

To start, suppose $f(x)$ and $g(x)$ are inverses. Let’s use what we know about the geometry of inverses to gain insight about how $g'(x)$ and $f'(x)$ are related.

The geometric derivation above concludes that if f and g are inverse functions, then the derivative of $g(x)$ at $x=a$ equals the reciprocal of the derivative of the function $f(x)$ at the point $\boldsymbol{x = g(a)}.$

Inverse Function theorem:
If two functions $f$ and $g$ are inverses, then
\[g'(x) = \dfrac{1}{f'(g(x))}\] Equivalently, since $g(x) = f^{-1}(x)$:
\[\big[f^{-1}\big]'(x) = \frac{1}{f’\Big(f^{-1}(x)\Big)}\]

The last line above can be difficult to parse and remember, due to the primes and inverse “–1” notation, and so we recommend remembering the formula using g and f instead.

The key insight is that to find the derivative of a function’s inverse $\left[f^{-1}\right]'(x)$, you can take the reciprocal of the derivative of the function $f$, and evaluate it at $f^{-1}$.

Tips iconThe most common student error is using the wrong input value on the right side of the equation above. Keeping the geometric picture from the derivation in mind (below) can help you easily avoid this mistake.
The curve y = g(x) = f^{-1}(x) and the tangent line at x = a. Also the curve y = f(x) and the tangent line at x = g(a). The slopes of these two tangent lines are reciprocals of each other.

Development of the Inverse Function theorem
using the Chain Rule

Let’s now use the Chain Rule to develop the same result.

We begin with a fundamental property of inverse functions: If g and f are inverse functions, then
\[f\Big(g(x)\Big)= x\] That is, if you input x into g, and then use that output $g(x)$ as input into g’s inverse function f, you get back out x again. This is the defining property of inverse functions.

Let’s now take the derivative of both sides of that equation, remembering to apply the Chain Rule to the left-hand side:
\big[f\Big(g(x)\Big)\big]’ &= [x]’ \\[8px] f’\Big(g(x)\Big) \cdot g'(x) &= 1 \\[8px] g'(x) &= \frac{1}{f’\Big(g(x)\Big)} \quad \blacktriangleleft
This of course is the same result as we found above geometrically. Notice in particular that, once again, we evaluate $f’$ on the right side of the equation at the input value $g(x).$ While this derivation is clearly more straightforward than the geometric one above, it doesn’t offer the same insight into why we evaluate $f’$ using that input value (which, we imagine, is the reason so very many students get that wrong on exams). Again, keeping the geometric picture above in mind will assure that you use the correct input value.

Examples and Scaffolded Practice

Let’s consider some Examples.

Example 1: Table of values
The table show input and output values for a smooth, continuous, one-to-one function $f(x),$ along with values of its derivative $f'(x).$
Find $\big[f^{-1}\big]'(1)$.


We use the Inverse Function theorem:
\[\big[f^{-1}\big]'(1) = \frac{1}{f’\Big(f^{-1}(1)\Big)}\] The crucial first move here is to find $f^{-1}(1)$ using the table. That is, we’re looking for the input x to f which gives an output $f(x)=1$. That value is $x=0$: since $f(0)=1$, we have $f^{-1}(1)=0$.

Next we look in the same row of the table to find
\[f’\Big(f^{-1}(1)\Big) = f'(0) = \frac{19}{12} \]

Then finally using the Inverse Function theorem,
\big[f^{-1}\big]'(1) &= \frac{1}{f’\Big(f^{-1}(1)\Big)}\\[8px] &= \dfrac{1}{19/12} \\[8px] &= \frac{12}{19} \quad \cmark


In the next Example we’ll use three different approaches to calculate the derivative of the inverse function of $f(x) = x^2.$ The real value is in showing how the various routes we now have available all lead to the same result, and you can see how the pieces fit together.

Scaffolded Problem #1

Using the same table as Example 1, find $[f^{-1}]'(3)$.

From the Inverse Function theorem we know
\[\big[f^{-1}\big]'(3) = \frac{1}{f’\Big(f^{-1}(3)\Big)}\]

Step 1. Find $f^{-1}(3).$

Show/Hide Step 1 Solution

From the table we see that $f(2) = 3,$ and so $f^{-1}(3) = 2$. This means we want to look at the row of the table for $x=2.$

Step 2. Use the value of $f’\Big(f^{-1}(3)\Big)$ to compute $[f^{-1}]'(3).$

Show/Hide Step 2 Solution

From the table, we see
\[f’\big(f^{-1}(3)\big) = f'(2) = \frac{21}{12}\] \begin{align*}
\left[f^{-1}\right]'(3) &= \frac{1}{f’\big(f^{-1}(3)\big)} \\[8px] &= \frac{1}{21/12} \\[8px] &= \frac{12}{21} \quad \cmark
Example 2: $f(x) = x^2$

Consider the function $f(x) = x^2$ for $x \ge 0.$ Find $\left[f^{-1}\right]'(4)$.

We’ll solve this using three approaches to illustrate various ways you can find the result.

Method 1. Use the explicit inverse function $g(x) = f^{-1}(x) = \sqrt{x}.$
This is the method we would have used before we developed the results on this screen. Since $f(x) = x^2,$ we know immediately that its inverse function is $g(x) = f^{-1}(x) = \sqrt{x}.$ We further know that $g'(x) = \dfrac{1}{2\sqrt{x}}$, and so
\[\left[f^{-1}\right]'(4) = g'(4) = \dfrac{1}{2 \sqrt{4}} = \dfrac{1}{4} \quad \cmark \]

Let’s see how the Inverse Function theorem leads to the same result.

Method 2. Use the Inverse Function theorem.
\[\big[f^{-1}\big]'(4) = \frac{1}{f’\Big(f^{-1}(4)\Big)}\] The input to $f’$ in the denominator is $f^{-1}(4),$ so let’s figure that out first: What input value of x (where $x \ge 0$) gives an output value of $f(x) = x^2 = 4$? We can answer “by inspection”: we look at the equation, and simply immediately know that $x = 2$: $f(2) = 4.$ Hence
\[f^{-1}(4) = 2 \quad \blacktriangleleft\]

Next, we find the derivative of f:
f(x) &= x^2 \\[8px] f'(x) &= 2x \quad \blacktriangleleft
Then finally
\big[f^{-1}\big]'(4) &= \frac{1}{f’\Big(f^{-1}(4)\Big)} \\[8px] &= \frac{1}{f'(2)} \\[8px] &= \frac{1}{2(2)} \quad [\text{since }f'(x) = 2x] \\[8px] &= \frac{1}{4} \quad \cmark
Method 3. Start with the defining relation for inverse functions, and use the Chain Rule.
Let’s say you didn’t immediately remember the Inverse Function theorem. (It happens!) If you’re given an expression for the function f, you can always proceed from the defining relation for inverse functions, $f^{-1}\big(f(x)\big) = x.$ We’ll use the notation $g(x) = f^{-1}(x)$ since it’s a little easier to write and track everything without the –1’s floating around:
f^{-1}\left(x^2 \right) = g\left(x^2\right) &= x \\[8px] \Big[g\left(x^2\right)\Big]’ &= [x]’ \\[8px] g’\left(x^2\right) \cdot (2x) &= 1 \quad [\text{[Remember the Chain Rule!]}] \\[8px] g’\left(x^2\right) &= \frac{1}{2x}
Notice that we actually just developed the Inverse Function theorem for the particular function $f(x) = x^2,$ which is what always happens with this method.
We still have to realize that since we have $g’\left(x^2\right)$ on the left-hand side of the equation and we want $g'(4),$ that means we have $x=2$:
g’\left(x^2\right) &= \frac{1}{2x} \\[8px] g'(4) &= \frac{1}{2(2)} \\[8px] &= \frac{1}{4} \quad \cmark

Let’s consider a third Example, now one where we don’t immediately know what the inverse function is, and so we instead move immediately to use the Inverse Function theorem.

Example 3: $p(x) = x^3 + 3x + 1$

Let $p(x) = x^3 + 3x + 1$. Use the Inverse Function theorem to find $[p^{-1}]'(5)$
Hint: Use inspection to determine $p^{-1}(5)$: What input value of x gives $p(x) = x^3 + 3x + 1 = 5$? Good choices to start guessing are often $x= 0 $ and $x = 1.$

The Inverse Function theorem says
\[\left[p^{-1}\right]'(5) = \frac{1}{p’\big(p^{-1}(5)\big)}\] To find $p^{-1}(5)$, we can do a little bit of guess work. Remember that we’re looking for x such that $x^3 + 3x + 1 = 5$.

Let’s try $x=0$:
\[p(0) = 0 + 0 + 1 = 1 \quad \xmark\] That’s not what we were looking for. Let’s next try $x=1$:
\[p(1) = 1 + 3 + 1 = 5 \quad \blacktriangleleft\] Ah, that’s it: $p(1) = 5.$ Therefore, $p^{-1}(5) = 1$.

Next, let’s find $p'(x)$:
p(x) &= x^3 + 3x + 1 \\[8px] p'(x) &= 3x^2 + 3
\[p'(1) = 3(1)^2 + 3 = 6 \quad \blacktriangleleft\] Now returning to the Inverse Function theorem:
\left[p^{-1}\right]'(5) &= \frac{1}{p'(p^{-1}(5))} \\[8px] &= \frac{1}{p'(1)} \\[8px] &= \frac{1}{6} \quad \cmark

Practice Problems

Time to practice!

Practice Problem #1
If $g(2)=3,$ $g(3) = 5$, $g'(2)=\dfrac{1}{6}$, and $g'(3) = \dfrac{1}{3},$ and $g'(5) = \dfrac{1}{2},$ calculate $\left[g^{-1}\right]'(3)$. \begin{array}{lllll} \text{(A) }\dfrac{1}{3} && \text{(B) }6 && \text{(C) }3 && \text{(D) }\dfrac{1}{2} && \text{(E) none of the above} \end{array}
Show/Hide Solution
\[ \left[ g^{-1} \right]'(3) = \dfrac{1}{g’\big( g^{-1}(3) \big)} \] Since we are told $g(2)=3,$ we know that $g^{-1}(3) = 2.$ We also know $g'(2)=\dfrac{1}{6}.$
\begin{align*} \left[ g^{-1} \right]'(3) &= \dfrac{1}{g’\left( g^{-1}(3) \right)} \\[8px] &= \dfrac{1}{g’\left( 2 \right)} \\[8px] &= \dfrac{1}{1/6} \\[8px] &= 6 \implies \; \text{ (B) } \; \cmark \end{align*}
[hide solution]
Practice Problem #2
value of x$f(x)$$f'(x)$
Given the values for $x$, $f$, and $f'$ from the table, find $ \left[ f^{-1} \right]'(0)$. \begin{array}{lllll} \text{(A) }\dfrac{1}{3} && \text{(B) }-1 && \text{(C) }1 && \text{(D) }2 && \text{(E) None of the above} \end{array}
Show/Hide Solution
We use the Inverse Function theorem: \[ \left[ f^{-1} \right]'(x) = \dfrac{1}{f’ \big( f^{-1}(x) \big)} \] $f^{-1}(0) = -1$ and $f'(-1) = 1$. \begin{align*} \left[ f^{-1} \right]'(0) &=\dfrac{1}{f’ \big( f^{-1}(0) \big)} = \dfrac{1}{f'(-1)} \\[8px] &= \dfrac{1}{1} \\[8px] &= 1 \implies \; \text{ (C) } \; \cmark \end{align*}
[hide solution]
Practice Problem #3
Given $f(x) = x^{3} + 3x + 1$ find $ \left[ f^{-1} \right]'(1) $ without explicitly finding $f^{-1}(x).$
[Hint: Determine $f^{-1}(1)$ by inspection. What input value of $x$ gives $f(x) = 1$?] \begin{array}{lllll} \text{(A) }\dfrac{1}{3} && \text{(B) }\dfrac{1}{6} && \text{(C) }3 && \text{(D) }6 && \text{(E) none of the above} \end{array}
Show/Hide Solution
We use the derivative of Inverse Function theorem:
\[ \left[ f^{-1} \right]'(1) = \dfrac{1}{f’ \big( f^{-1}(1) \big)} \] We first must determine $f^{-1}(1),$ which we do “by inspection”: Since \[f(x) = x^{3} + 3x + 1\] we see that the input value $x=0$ gives us $f(0) = 1.$ Hence $f^{-1}(1) = 0.$

Then we find \begin{align*} f'(x) &= 3x^2 + 3 \\[8px] f'(0) &= 3 \blacktriangleleft \end{align*} So $f’\big( f^{-1}(1)\big) = f'(0) = 3.$

Finally, \begin{align*} \left[ f^{-1} \right]'(1) &= \dfrac{1}{f’\big( f^{-1}(1)\big)} \\[8px] &= \dfrac{1}{3} \implies \; \text{ (A) } \; \cmark \end{align*}
[hide solution]
Practice Problem #4
Graph of the inverse function f^{-1}(x), which consists of two line segments. The first starts at (0, -4) and ends at (4, 4). The second starts at (4, 4) and ends at (8, 6).

The graph shows the inverse of a particular function $f$. Find $f'(0)$.
\begin{array}{lllll} \text{(A) }\dfrac{1}{2} && \text{(B) }2 && \text{(C) }0 && \text{(D) }-2 && \text{(E) }-\dfrac{1}{2}

Show/Hide Solution
We start with the Inverse Function theorem, noting that we want $f'(0)$ so let’s solve for $f’$:
\begin{align*} \left[ f^{-1} \right]'(x) &= \dfrac{1}{f’\left( f^{-1}(x) \right)} \\[8px] f’\left( f^{-1} (x) \right) &= \dfrac{1}{\left[ f^{-1} \right]'(x)} \\[8px] \end{align*} From the graph, we see $f^{-1}(x) = 0$ when $x=2$. Hence, \begin{align*} f'(0) &= \dfrac{1}{\left[ f^{-1} \right]'(2)} \\[8px] \end{align*} From the graph, we see the slope of $f^{-1}(x)$ at $x=2$ is $2$: $[f^{-1}]'(2) = 2.$ Subsituting this value gives \begin{align*} f'(0) &= \dfrac{1}{\left[ f^{-1} \right]'(2)} \\[8px] &= \dfrac{1}{2} \implies \; \text{ (A) } \; \cmark \end{align*}
[hide solution]
Practice Problem #5
Consider the function $f(x)= \sin (2x - \pi)$ on the interval $0 \le x \lt \pi$. Find the value of $\left[ f^{-1} \right]'(0)$. \begin{array}{lllll} \text{(A) }0 && \text{(B) }\pi && \text{(C) } \dfrac{\pi}{2} && \text{(D) }\dfrac{1}{2} && \text{(E) none of the above} \end{array}
Show/Hide Solution
\[ \left[ f^{-1} \right]'(0) = \dfrac{1}{f’ \big( f^{-1}(0) \big)} \] We first determine $f^{-1}(0)$ given $f(x)= \sin (2x – \pi)$ on the interval $0 \le x \lt \pi$.

To do so, make use of your trig knowledge: We want the input value of x that produces $f(x) = 0.$ We know that on the interval $0 \le \theta \lt \pi,$ the value of $\theta$ that gives $\sin (\theta) = 0$ is $\theta = 0.$ (You could also use your calculator to find $\sin^{-1}(0) = 0.$) Hence \begin{align*} 2x – \pi &= 0 \\[8px] x &= \dfrac{\pi}{2} \; \blacktriangleleft \end{align*} Then, \begin{align*} f’\left( \dfrac{\pi}{2} \right) &= 2\left[ \cos\Big( 2\left( \frac{\pi}{2} \right) – \pi \Big) \right] \\[8px] &= 2 \cos (\pi – \pi) \\[8px] &= 2 \cos (0) \\[8px] &= 2 \\[8px] \end{align*} That is, $f’ \big( f^{-1}(0) \big) = f’\left(\dfrac{\pi}{2} \right) = 2.$

Finally, we can apply the Inverse Function Theorem.
\begin{align*} \left[ f^{-1} \right]'(0) &= \dfrac{1}{f’ \big( f^{-1}(0) \big)} \\[8px] &= \dfrac{1}{2} \implies \; \text{ (D) } \; \cmark \end{align*}
[hide solution]

The Upshot

  1. The Inverse Function theorem states that if two functions $f$ and $g$ are inverses, then
    \[g'(x) = \dfrac{1}{f'(g(x))}\] Equivalently, since $g(x) = f^{-1}(x)$:
    \[\big[f^{-1}\big]'(x) = \frac{1}{f’\Big(f^{-1}(x)\Big)}\]

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