Inverse functions are functions that “reverse” each other.

We consider a function \(f\left( x \right)\), which is strictly monotonic on an interval \(\left( {a,b} \right)\). If there exists a point \({x_0}\) in this interval such that \(f’\left( {{x_0}} \right) \ne 0\), then the inverse function \(x = \varphi \left( y \right)\) is also differentiable at \({y_0} = f\left( {{x_0}} \right)\) and its derivative is given by

\[\varphi’\left( {{y_0}} \right) = \frac{1}{{f’\left( {{x_0}} \right)}}.\]

Let us prove this theorem (called the inverse function theorem).

Suppose that the variable \(y\) gets an increment \(\Delta y \ne 0\) at the point \({y_0}.\) The corresponding increment of the variable \(x\) at the point \({x_0}\) is denoted by \(\Delta x\), where \(\Delta x \ne 0\) due to the strict monotonicity of \(y = f\left( x \right)\). The ratio of the increments is written as

\[\frac{{\Delta x}}{{\Delta y}} = \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}}.\]

Suppose that \(\Delta y \to 0\). Then \(\Delta x \to 0\), since the inverse function \(x = \varphi \left( y \right)\) is continuous at \({y_0}\). In the limit when \(\Delta x \to 0\), the right side of the relationship becomes

\[

{\lim\limits_{\Delta x \to 0} \frac{1}{{\frac{{\Delta y}}{{\Delta x}}}} = \frac{1}{{\lim\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}}} }

= {\frac{1}{{f’\left( {{x_0}} \right)}}.}

\]

In this case, the left hand side also approaches a limit, which by definition is equal to the derivative of the inverse function:

\[\lim\limits_{\Delta y \to 0} \frac{{\Delta x}}{{\Delta y}} = \varphi’\left( {{y_0}} \right).\]

Thus,

\[\varphi’\left( {{y_0}} \right) = \frac{1}{{f’\left( {{x_0}} \right)}},\]

that is the derivative of the inverse function is the inverse of the derivative of the original function.

In the examples below, find the derivative of the function \(y = f\left( x \right)\) using the derivative of the inverse function \(x = \varphi \left( y \right).\)

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

\[y = \sqrt[\large n\normalsize]{x}\]### Example 2

\[y = \arcsin x\]### Example 3

\[y = \ln x\]### Example 4

\[y = \sqrt[\large 3\normalsize]{{x + 1}}\]### Example 5

\[y = \arccos \left( {1 – 2x} \right)\]### Example 6

\[y = \sqrt {1 + \sqrt x } \]### Example 7

\[y = \arctan \frac{1}{x}\]### Example 8

\[y = \sqrt x \]### Example 9

\[y = 2x + 4\]### Example 10

Given the function \(y = {x^5} + 2{x^3} + 3x\). Find the derivative of the inverse function at \(x = 1.\)### Example 11

Given the function \(y = {x^2} – x\). Find the derivative of the inverse function at \(x = 1.\)### Example 12

Given the function \(y = {e^x} + 2x + 1\). Find the derivative of the inverse function at \(x = 0.\)### Example 13

Find the derivative of the inverse function at \(x = 1\) for the function \(y = \sin \left( {x – 1} \right) + {x^2}.\)### Example 14

Find the derivative of the inverse function of \(y = {x^2} + 2\ln x\) and calculate its value at \(x = 1.\)### Example 15

Find the derivative of the inverse function for \(y = {x^3} – 3x\) and calculate its value at \(x = -2.\)### Example 16

Find the derivative of the inverse function for \(y = 2{x^3} – 1\) and calculate its value at \(x = 2.\)### Example 17

\[y = {\log _2}\left( {\frac{x}{3}} \right)\]### Example 18

Find the derivative of \(y = \text{arcsec }x\) at \(x = \sqrt 2.\)### Example 19

Find the derivative of the inverse function for \(y = x \cdot {3^x}\) provided \(x \gt 0.\)### Example 20

Find the derivative of the inverse hyperbolic sine function \(y = \text{arcsinh } x.\)### Example 1.

\[y = \sqrt[\large n\normalsize]{x}\]Solution.

We first determine the inverse function for the given function \(y = \sqrt[\large n\normalsize]{x}\). To do this, we express the variable \(x\) in terms of \(y:\)

\[

{y = f\left( x \right) = \sqrt[\large n\normalsize]{x},\;\;}\Rightarrow

{{y^n} = {\left( {\sqrt[\large n\normalsize]{x}} \right)^n},\;\;}\Rightarrow

{x = \varphi \left( y \right) = {y^n}.}

\]

By the inverse function theorem, we can write:

\[

{{\left( {\sqrt[\large n\normalsize]{x}} \right)^\prime } = f’\left( x \right) }

= {\frac{1}{{\varphi’\left( y \right)}} }

= {\frac{1}{{{{\left( {{y^n}} \right)}^\prime }}} }

= {\frac{1}{{n{y^{n – 1}}}}.}

\]

Now we substitute \(y = \sqrt[\large n\normalsize]{x}\) instead of \(y.\) As a result, we obtain an expression for the derivative of the given function:

\[

{{\left( {\sqrt[\large n\normalsize]{x}} \right)^\prime } = \frac{1}{{n{y^{n – 1}}}} }

= {\frac{1}{{n{{\left( {\sqrt[\large n\normalsize]{x}} \right)}^{n – 1}}}} }

= {\frac{1}{{n\sqrt[\large n\normalsize]{{{x^{n – 1}}}}}}\;\;\;}\kern-0.3pt{\left( {x \gt 0} \right).}

\]

### Example 2.

\[y = \arcsin x\]Solution.

The arcsine function is the is the inverse of the sine function. Therefore \(x = \varphi \left( y \right) \) \(= \sin y.\) Then the derivative of \(\arcsin x\) is

\[

{{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) }

= {\frac{1}{{\varphi’\left( y \right)}} }

= {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }

= {\frac{1}{{\cos y}} }

= {\frac{1}{{\sqrt {1 – {{\sin }^2}y} }} }

= {\frac{1}{{\sqrt {1 – {{\sin }^2}\left( {\arcsin x} \right)} }} }

= {\frac{1}{{\sqrt {1 – {x^2}} }},}

\]

where \(-1 \lt x \lt 1.\)

### Example 3.

\[y = \ln x\]Solution.

The natural logarithm and the exponential function are mutually inverse functions. Therefore, \(x = \varphi \left( y \right) = {e^y}\), where \(x \gt 0\), \(y \in \mathbb{R}\). The derivative of the natural logarithm is easy to calculate through the derivative of the exponential function:

\[

{{\left( {\ln x} \right)^\prime } = f’\left( x \right) }

= {\frac{1}{{\varphi’\left( y \right)}} }

= {\frac{1}{{{{\left( {{e^y}} \right)}^\prime }}} }

= {\frac{1}{{{e^y}}} }

= {\frac{1}{{{e^{\ln x}}}} }

= {\frac{1}{x}}

\]

Here we have used a logarithmic identity, according to which

\[{e^{\ln x}} = x.\]