Derivative of Inverse Function

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    Derivatives of Inverse Functions heartbeat electrocardiogram StudySmarterThe screen of an electrocardiogram

    The information gathered from a device is first transformed using a function, so it can be processed. After this is done it is necessary to undo the transformation using an inverse function. This processing can be finding a derivative, and sometimes it is even possible to work on the derivative of the inverse function itself! In this article, we will take a look at how this is done.

    Rule for the Derivative of Inverse Functions

    If you know the derivative of a function you can find the derivative of its inverse without using the definition of a derivative. Here is how you can do it.

    Let \( f(x) \) be an invertible and differentiable function, and let \( f^{-1}(x) \) be its inverse. If \( f^{-1}(x) \) is differentiable, its derivative is given by the following formula:

    $$\left( f^{-1} \right)' (x) = \frac{1}{f'\left( f^{-1}(x) \right)}.$$

    This means that you need to find the derivative of \( f(x) \) and find its composition with \( f^{-1}(x). \) Assuming that \( f^{-1}(x) \) is known, this procedure can be summarized in the following steps:

    1. Find the derivative of \( f(x) \), that is, find \( f'(x). \)

    2. Find the composition \( f' \left( f^{-1}(x) \right). \)

    3. Take the reciprocal of \( f' \left( f^{-1}(x) \right). \)

    This is better understood by looking at some examples.

    Derivatives of Inverse Functions Examples

    There is a wide variety of invertible functions that we can differentiate, so let's take a look at some examples.

    Derivatives of Irrational Functions

    Square root functions and quadratic functions are inverses of each other. You can find the derivative of a quadratic function by using the Power Rule, and then, use this result to find the derivative of a square root function.

    Consider the function \( f(x)=x^2. \) Its inverse is the square root function \( f^{-1}(x)=\sqrt{x}. \) Find the derivative of the square root function.

    Answer:

    1. Find the derivative of \( f(x).\)

    To use the derivative of an inverse function formula you first need to find the derivative of \( f(x). \) In this case you can use The Power Rule, so

    $$f'(x)=2x.$$

    2. Find the composition \( f' \left( f^{-1}(x) \right). \)

    You can find the composition by using \( f^{-1}(x) \) as the input of \( f'(x). \) Take the derivative

    $$f'(x)=2x,$$

    and replace \(x \) for \(\sqrt{x},\) which gives you

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

    3. Take the reciprocal of \( f' \left( f^{-1}(x) \right). \)

    The final step is to take the reciprocal of the expression you just got in the last step, so

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

    Let's now take a look at an example of a cubic function.

    Consider the function \( g(x)=x^3. \) Its inverse is the cubic root function \( g^{-1}(x)= \sqrt[3]{x}.\) Find the derivative of the cube root function.

    Answer:

    You can find the derivative of the cubic root function using a similar procedure.

    1. Find the derivative of \( g(x).\)

    You can use the Power Rule to find the derivative of \( g(x), \)

    $$g'(x)=3x^2.$$

    2. Find the composition \( g' \left( g^{-1}(x) \right). \)

    Next, you need to find the composition of the above derivative with the cubic root function, so

    $$\begin{align}g' \left( g^{-1}(x) \right) &= 3\left( \sqrt[3]{x}\right)^2 \\[0.5em] &=3x^{^2/_3}. \end{align}$$

    3. Take the reciprocal of \( g' \left( g^{-1}(x) \right). \)

    Finally, take the reciprocal of the expression you got in the previous step, which can be rewritten using properties of exponents

    $$\begin{align}\left( g^{-1}\right)' (x) &= \frac{1}{3x^{^2/_3}} \\[0.5em] &= \frac{1}{3}x^{^{-2}/_3}.\end{align}$$

    Derivatives of Logarithmic Functions

    Even though you can find the derivative of logarithmic functions using the definition of a derivative, you can also use the fact that the logarithmic function is the inverse of the exponential function.

    Let \( f(x)=e^x.\) The inverse of the exponential function is the natural logarithm function, that is \( f^{-1}(x)=\ln{x}.\) We Find the derivative of the natural logarithmic function.

    Answer:

    1. Find the derivative of \( f(x).\)

    Begin by finding the derivative of the exponential function, which is itself, that is

    $$f'(x)=e^x.$$

    2. Find the composition \( f' \left( f^{-1}(x) \right). \)

    Having the derivative of an exponential function be itself makes the composition rather easy, as a function composed with its inverse is equal to \( x, \) that is

    $$\begin{align}f' \left( f^{-1}(x) \right) &= e^{\ln{x}} \\[0.5em] &= x. \end{align}$$

    3. Take the reciprocal of \( f' \left( f^{-1}(x) \right). \)

    Finally, take the reciprocal of the expression from the above step to get the derivative of the natural logarithmic function

    $$\left( f^{-1} \right)'(x)=\frac{1}{x}.$$

    This procedure is an excellent alternative to finding the derivative of the natural logarithmic function using the definition of a derivative!

    Common Mistakes When Finding the Derivative of an Inverse Function

    There are two common mistakes when finding the derivative of an inverse function.

    • Doing the composition in the wrong order.

    • Forgetting to take the reciprocal of the composition.

    Let's take a look at each.

    Doing the composition in the wrong order

    One common mistake is getting the composition in the wrong order. Remember that, in general

    $$f'\left( f^{-1}(x)\right) \neq f^{-1}\left( f'(x) \right).$$

    Let's see an example of a composition done in the wrong order.

    Let's go back to the quadratic function example \( f(x)=x^2. \) You found Its derivative to be \(f'(x)=2x\) and its inverse is \( f^{-1}(x)=\sqrt{x}.\) What happens if you do the composition in the wrong order? You would have

    $$f^{-1}\left( f'(x) \right) = \sqrt{2x},$$

    which has a \(2\) inside the square root, rather than outside it, as we found before. You don't even need to take the reciprocal, this will give you already a different result!

    Forgetting to take the reciprocal of the composition

    Another common mistake is forgetting to take the reciprocal after finding the composition. That is

    $$\left( f^{-1} \right) ' (x) \neq f'\left( f^{-1}(x)\right),$$

    so always remember to take the reciprocal of the composition

    $$\left( f^{-1} \right) ' (x) = \frac{1}{f'\left( f^{-1}(x)\right)}.$$

    Derivatives of Inverse Trigonometric Functions

    The inverse functions of trigonometric functions are usually just called Inverse trigonometric functions. They are also known as arcus functions. You can use the Derivative of an Inverse Function Formula to find the derivatives of inverse trigonometric functions!

    Let \( f(x)=\sin{x}.\) The inverse sine function is \( f^{-1}(x)=\arcsin{x}.\)

    Find the derivative of the inverse sine function.

    Answer:

    1. Find the derivative of \( f(x).\)

    First, you need the derivative of the sine function (take a look at our Derivatives of Trigonometric functions if you need a refresher)

    $$f'(x)=\cos{x}.$$

    2. Find the composition \( f' \left( f^{-1}(x) \right). \)

    If you try to do the composition right now, it would be a little tricky. You can instead use the Pythagorean identity

    $$\sin^2{x}+\cos^2{x}=1$$

    to rewrite \( f'(x) \) in terms of the sine function, that is

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

    This way, as you compose the sine and the inverse sine you will get \( x,\) so the composition is given by

    $$\begin{align} f' \left( f^{-1}(x) \right) &= \sqrt{1-\left( \sin{ \left( \arcsin{x} \right) } \right)^2} \\[0.5em]&= \sqrt{1-x^2}.\end{align}$$

    3. Take the reciprocal of \( f' \left( f^{-1}(x) \right). \)

    As usual, the last step is to take the reciprocal of the above expression

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

    giving you the derivative of the inverse sine function.

    A similar procedure can be used to find the derivative of the inverse tangent function. Let's take a look at the following example.

    Let \( g(x)=\tan{x}.\) The inverse tangent function is \( g^{-1}(x)=\arctan{x}.\) Find the derivative of the inverse tangent function.

    Answer:

    1. Find the derivative of \( g(x).\)

    Begin by finding the derivative of the tangent function, that is

    $$g'(x)=\sec^2{x}.$$

    2. Find the composition \( g' \left( g^{-1}(x) \right). \)

    Just like in the derivative of the inverse sine, you need to write this in terms of the tangent function. This time the Pythagorean identity

    $$\tan^2{x}+1=\sec^2{x}$$

    does the job. This way, the derivative of the tangent function is

    $$g'(x)=\tan^2{x}+1.$$

    This makes the composition fairly more simple, so

    $$\begin{align} g' \left( f^{-1}(x) \right) &= \left( \tan{ \left( \arctan{x} \right) } \right)^2+1 \\[0.5em]&= x^2+1,\end{align}$$

    3. Take the reciprocal of \( f' \left( f^{-1}(x) \right). \)

    Finally, you take the reciprocal of the expression obtained in the previous step to get

    $$\left( g^{-1} \right)'(x)=\frac{1}{x^2+1}.$$

    Derivative of an inverse function from a table

    Functions can be represented in many ways. Take for instance a table of values.

    \( x \)\( f(x)=x^2 \)
    00
    11
    24
    39
    416

    You can use the above table to find the slope of a line secant to the function by taking two points an applying the slope formula

    \[ m=\frac{y_2-y_1}{x_2-x_1}\]

    as shown in the following image.

    Derivatives of Inverse Functions secant line StudySmarterFig. 1. Graph of a line secant to the function at two points.

    In the above image the points \( (1,f(1)) \) and \( (2,f(2)) \) are being used for the secant line, so its slope is

    \[ \begin{align} m &= \frac{f(2)-f(1)}{2-1} \\ &= \frac{4-1}{1} \\ &= 3. \end{align} \]

    By switching the values of the table you can get the inverse of the original function, in this case you will get the square root function.

    \( x \)\( f^{-1}(x) \)
    00
    11
    42
    93
    164

    You can also find a line secant to the square root function by using two points. The first point can stay the same as it is shared amongst both functions. For the second point rather than using \( (2,f^{-1}(2)) \) you have compose the function, that is, you need to use \( (f(2),f^{-1}(f(2))).\) Therefore the corresponding secant uses \( (1,f^{-1}(1)) \) and \( (4,f^{-1}(4)) \) instead as shown in the following image.

    Derivatives of Inverse Functions secant line StudySmarterFig. 2. Graph of a line secant to the inverse function at two points.

    This time the points \( (1,f^{-1}(1)) \) and \( (4,f^{-1}(4)) \) are being used for the secant line, so its slope is

    \[ \begin{align} m &= \frac{f^{-1}(4)-f^{-1}(1)}{4-1} \\ &= \frac{2-1}{3} \\ &= \frac{1}{3}. \end{align} \]

    Note that the slope of the secant of the inverse function is the reciprocal of the slope of the line secant to the original function. Furthermore, you need to compose the function and the inverse to find the above slope. Do these steps sound familiar?

    The slope of secant lines are related to derivatives by means of a limit. The above reasoning keeps working as you take smaller intervals, connecting to the formula of the derivative of an inverse function.

    Proof of the Derivative of an Inverse Function Formula

    The proof of the Derivative of an Inverse Function uses the fact that the composition of a function and its inverse is equal to the identity function, that is

    $$f\left(f^{-1}(x)\right)=x.$$

    Next, differentiate both sides of the equation. You use the Power Rule to differentiate the right-hand side of the equation, so

    $$\frac{\mathrm{d}}{\mathrm{d}x} f\left(f^{-1}(x)\right) = 1.$$

    The left-hand side of the equation can be differentiated using the Chain Rule, giving you

    $$f'\left(f^{-1}(x)\right) \left( f^{-1} \right) ' (x)=1,$$

    and finally, you can isolate the derivative of the inverse function

    $$\left( f^{-1} \right)'(x)=\frac{1}{f'\left( f^{-1}(x) \right)}.$$

    Derivatives of Inverse Functions - Key takeaways

    • The formula to find the derivative of the inverse of a function is given as follows: $$\left( f^{-1} \right)'(x)=\frac{1}{f'\left( f^{-1}(x) \right)}.$$
    • The process of finding the derivative of an inverse function can be summarized in the following steps:
      • Find the derivative of \( f(x). \)
      • Find the composition \( f' \left( f^{-1}(x) \right).\)
      • Take the reciprocal of \( f' \left( f^{-1}(x) \right).\)
    • There are two common mistakes when finding the derivative of the inverse of a function:
      • Doing the composition in the wrong order.
      • Forgetting to take the reciprocal of the composition.
    Frequently Asked Questions about Derivative of Inverse Function

    What is an inverse function differentiation? 

    It is the process of finding the derivative of the inverse of a function.

    How do you prove the differentiation of an inverse function? 

    Compose the function and its inverse to form the identity function. From there, perform implicit differentiation and use the chain rule. Finally, isolate the derivative of the inverse function.

    What are the steps for inverse function differentiation? 

    1. Find the derivative of the original function.
    2. Find the composition of the derivative of the original function and the inverse function.
    3. Take the reciprocal of the composition.

    What are the steps and methods in inverse function derivation?

    To obtain the derivative of an inverse function you only need to obtain the in verse function and then apply the normal rules of derivation.

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    Test your knowledge with multiple choice flashcards

    Let \(f(x)\) be a differentiable function. To find the derivative of its inverse, you also need it to be a(n) _______ function.

    The rule for finding the derivative of the inverse of a function is:

    The following expression denotes the derivative of the inverse of a function:

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