Derivatives of Inverse Functions

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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

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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 \(y:\)

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