# Derivatives of Inverse Functions

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

$y = \sqrt[n]{x}$

### Example 2

$y = \arcsin x$

### Example 3

$y = \ln x$

### Example 4

$y = \sqrt[3]{{x + 1}}$

### Example 5

$y = \arccos \left( {1 - 2x} \right)$

### Example 6

$y = \sqrt {1 + \sqrt x }$

### Example 1.

$y = \sqrt[n]{x}$

Solution.

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

$y = f\left( x \right) = \sqrt[n]{x},\;\; \Rightarrow {y^n} = {\left( {\sqrt[n]{x}} \right)^n},\;\; \Rightarrow x = \varphi \left( y \right) = {y^n}.$

By the inverse function theorem, we can write:

$\left( {\sqrt[n]{x}} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {{y^n}} \right)}^\prime }}} = \frac{1}{{n{y^{n - 1}}}}.$

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

$\left( {\sqrt[n]{x}} \right)^\prime = \frac{1}{{n{y^{n - 1}}}} = \frac{1}{{n{{\left( {\sqrt[n]{x}} \right)}^{n - 1}}}} = \frac{1}{{n\sqrt[n]{{{x^{n - 1}}}}}}\;\;\;\left( {x \gt 0} \right).$

### Example 2.

$y = \arcsin x$

Solution.

The arcsine function 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 the logarithmic identity

${e^{\ln x}} = x.$

### Example 4.

$y = \sqrt[3]{{x + 1}}$

Solution.

We first find the inverse function $$x = \varphi \left( y \right)$$ for the given function $$y = f\left( x \right)$$ which is monotonically increasing for any $$x \in \mathbb{R}$$. Express $$x$$ in terms of $$y:$$

$y = \sqrt[3]{{x + 1}},\;\; \Rightarrow {y^3} = x + 1,\;\; \Rightarrow x = {y^3} - 1.$

Now find the derivative $$f'\left( x \right):$$

$\left( {\sqrt[3]{{x + 1}}} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {{y^3} - 1} \right)}^\prime }}} = \frac{1}{{3{y^2}}} = \frac{1}{{3\sqrt[3]{{{{\left( {x + 1} \right)}^2}}}}}\;\;\left( {x \ne 1} \right).$

### Example 5.

$y = \arccos \left( {1 - 2x} \right)$

Solution.

The arccosine function is defined and monotonic on the interval $$\left[ { - 1,1} \right]$$. Consequently, the domain of the original function has the form:

$- 1 \le 1 - 2x \le 1,\;\; \Rightarrow - 2 \le - 2x \le 0,\;\; \Rightarrow 0 \le x \le 1.$

Write the inverse function $$x = \varphi \left( y \right):$$

$y = \arccos \left( {1 - 2x} \right),\;\; \Rightarrow 1 - 2x = \cos y,\;\; \Rightarrow 2x = 1 - \cos y,\;\; \Rightarrow x = \frac{1}{2} - \frac{1}{2}\cos y.$

Calculate the derivative of the original function through the derivative of the inverse function:

$\require{cancel} \left( {\arccos \left( {1 - 2x} \right)} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\frac{1}{2} - \frac{1}{2}\cos y} \right)}^\prime }}} = \frac{1}{{\frac{1}{2}\sin y}} = \frac{2}{{\sin y}} = \frac{2}{{\sqrt {1 - {\cos^2}y} }} = \frac{2}{{\sqrt {1 - {\cos^2}\left( {\arccos \left( {1 - 2x} \right)} \right)} }} = \frac{2}{{\sqrt {1 - {{\left( {1 - 2x} \right)}^2}} }} = \frac{2}{{\sqrt {1 - \left( {1 - 4x + 4{x^2}} \right)} }} = \frac{2}{{\sqrt {\cancel{1} - \cancel{1} + 4x - 4{x^2}} }} = \frac{\cancel{2}}{{\cancel{2}\sqrt {x - {x^2}} }} = \frac{1}{{\sqrt {x - {x^2}} }}.$

Note that the derivative is not defined at the boundary points $$x = 0$$ and $$x = 1$$ of the domain of the function $$y = f\left( x \right).$$

### Example 6.

$y = \sqrt {1 + \sqrt x }$

Solution.

This function is defined and monotonically increasing for $$x \gt 0$$. Therefore we can construct the inverse function on this interval. Express $$x$$ in terms of $$y:$$

$y = \sqrt {1 + \sqrt x } ,\;\; \Rightarrow {y^2} = 1 + \sqrt x ,\;\; \Rightarrow \sqrt x = {y^2} - 1,\;\; \Rightarrow x = {\left( {{y^2} - 1} \right)^2}.$

Now we define the derivative of the given function $$y = f\left( x \right)$$ using the inverse function theorem:

$\left( {\sqrt {1 + \sqrt x } } \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left[ {{{\left( {{y^2} - 1} \right)}^2}} \right]}^\prime }}} = \frac{1}{{2\left( {{y^2} - 1} \right) \cdot {{\left( {{y^2} - 1} \right)}^\prime }}} = \frac{1}{{2\left( {{y^2} - 1} \right) \cdot 2y}} = \frac{1}{{4y\left( {{y^2} - 1} \right)}}.$

Substitute the expression for the original function instead of $$y:$$

$\left( {\sqrt {1 + \sqrt x } } \right)^\prime = \frac{1}{{4y\left( {{y^2} - 1} \right)}} = \frac{1}{{4\sqrt {1 + \sqrt x } \left( {{{\left( {\sqrt {1 + \sqrt x } } \right)}^2} - 1} \right)}} = \frac{1}{{4\sqrt {1 + \sqrt x } \left( {\cancel{1} + \sqrt x - \cancel{1}} \right)}} = \frac{1}{{4\sqrt x \sqrt {1 + \sqrt x } }}\;\;\left( {x \gt 0} \right).$

See more problems on Page 2.