# Derivatives of Inverse Functions

## Solved Problems

### Example 7.

$y = \arctan \frac{1}{x}$

Solution.

The inverse function for this function is as follows:

$y = \arctan \frac{1}{x},\;\; \Rightarrow \frac{1}{x} = \tan y,\;\; \Rightarrow x = \frac{1}{{\tan y}},\;\;\text{where}\;\;x \ne 0.$

Find the derivative of the original function $$y = f\left( x \right):$$

$\left( {\arctan \frac{1}{x}} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\frac{1}{{\tan y}}} \right)}^\prime }}} = \frac{1}{{\left( { - \frac{1}{{{{\tan }^2}y}}} \right) \cdot \frac{1}{{{{\cos }^2}y}}}} = - \frac{{{{\tan }^2}y}}{{\frac{1}{{{{\cos }^2}y}}}}.$

Use the trigonometric identity

$\frac{1}{{{{\cos }^2}y}} = 1 + {\tan ^2}y.$

Then

$\left( {\arctan \frac{1}{x}} \right)^\prime = - \frac{{{{\tan }^2}y}}{{\frac{1}{{{{\cos }^2}y}}}} = - \frac{{{{\tan }^2}y}}{{1 + {{\tan }^2}y}} = - \frac{{{{\tan }^2}\left( {\arctan \frac{1}{x}} \right)}}{{1 + {{\tan }^2}\left( {\arctan \frac{1}{x}} \right)}} = - \frac{{{{\left( {\frac{1}{x}} \right)}^2}}}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} = - \frac{{\frac{1}{{{x^2}}}}}{{\frac{{{x^2} + 1}}{{{x^2}}}}} = - \frac{1}{{1 + {x^2}}}.$

It can be seen that the derivative of the function $$y = \arctan {\frac{1}{x}}$$ differs only in the sign of the derivative of the function $$y = \arctan x$$.

### Example 8.

$y = \sqrt x$

Solution.

This function is the inverse of the quadratic function $$x = \varphi \left( y \right) = {y^2}.$$ Therefore, its derivative is given by

$\left({\sqrt x}\right)' = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {{y^2}} \right)}^\prime }}} = \frac{1}{{2y}} = \frac{1}{{2\sqrt x }}\;\;\left( {x \gt 0} \right).$

### Example 9.

$y = 2x + 4$

Solution.

Write the function $$x = \varphi \left( y \right)$$ which is the inverse of the given function $$y = f\left( x \right):$$

$y = 2x + 4,\;\; \Rightarrow 2x = y - 4,\;\; \Rightarrow x = \frac{y}{2} - 2.$

Then the derivative $$f'\left( x \right)$$ has the following form:

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

### Example 10.

Given the function $$y = {x^5} + 2{x^3} + 3x$$. Find the derivative of the inverse function at $$x = 1.$$

Solution.

In this example, the finding common expression for the inverse function and its derivative would be too cumbersome. Therefore we calculate the derivative of the original function and then find the reciprocal.

$y' = f'\left( x \right) = \left( {{x^5} + 2{x^3} + 3x} \right)^\prime = 5{x^4} + 6{x^2} + 3.$

The value of the derivative $$f'\left( x \right)$$ at $$x = 1$$ is equal to

$f'\left( {x = 1} \right) = 5 \cdot {1^4} + 6 \cdot {1^2} + 3 = 14.$

The function itself at the point $$x = 1$$ takes the value

$y\left( {x = 1} \right) = {1^5} + 2 \cdot {1^3} + 3 \cdot 1 = 6.$

By the inverse function theorem, we obtain:

$\varphi'\left( {y = 6} \right) = \frac{1}{{f'\left( {x = 1} \right)}} = \frac{1}{{14}}.$

### Example 11.

Given the function $$y = {x^2} - x$$. Find the derivative of the inverse function at $$x = 1.$$

Solution.

Determine the derivative of the original function:

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

Note that the point $$x = {\frac{1}{2}}$$ separates the regions of decreasing ($$x \lt {\frac{1}{2}}$$) and increasing ($$x \gt {\frac{1}{2}}$$) of the original function. For each of the intervals there is its own branch of the inverse function, which are denoted as $${\varphi _1}\left( y \right)$$ and $${\varphi _2}\left( y \right).$$ The expressions for these functions can be obtained in explicit form by solving the equation $$y = f\left( x \right)$$ with respect to $$x:$$

$y = {x^2} - x,\;\; \Rightarrow {x^2} - x - y = 0,\;\; \Rightarrow D = 1 + 4y,\;\; \Rightarrow {\varphi _{1,2}}\left( y \right) = \frac{{1 \pm \sqrt {1 + 4y} }}{2}.$

The derivative of the inverse function is defined by the formula

$\varphi'\left( y \right) = \frac{1}{{f'\left( x \right)}} = \frac{1}{{2x - 1}}\;\;\;\left( {x \ne \frac{1}{2}} \right).$

Substituting the explicit expressions for $$x = {\varphi _1}\left( y \right)$$ and $$x = {\varphi _2}\left( y \right)$$ for both branches of the inverse function, we have:

${\varphi _1}^\prime \left( y \right) = \frac{1}{{2 \left( {\frac{{1 + \sqrt {1 + 4y} }}{2}} \right) - 1}} = \frac{1}{{\cancel{1} + \sqrt {1 + 4y} - \cancel{1}}} = \frac{1}{{\sqrt {1 + 4y} }},$
${\varphi _2}^\prime \left( y \right) = \frac{1}{{2 \left( {\frac{{1 - \sqrt {1 + 4y} }}{2}} \right) - 1}} = \frac{1}{{\cancel{1} - \sqrt {1 + 4y} - \cancel{1}}} = -\frac{1}{{\sqrt {1 + 4y} }}.$

The point $$x = 1$$ corresponds to the value $$y = {1^2} - 1 = 0$$ and is located on the branch $${\varphi _2}\left( y \right)$$ of the inverse function. The derivative of the inverse function at this point is given by

${\varphi _2}^\prime \left( {x = 1} \right) = {\varphi _2}^\prime \left( {y = 0} \right) = - \frac{1}{{\sqrt {1 + 4 \cdot 0} }} = - 1.$

### Example 12.

Given the function $$y = {e^x} + 2x + 1$$. Find the derivative of the inverse function at $$x = 0.$$

Solution.

When $$x = 0,$$ the given function takes the value

$y\left( {x = 0} \right) = {e^0} + 2 \cdot 0 + 1 = 2.$

The derivative of the function $$y = f\left( x \right)$$ and its value at the point $$x = 0$$ are equal:

$y' = f'\left( x \right) = \left( {{e^x} + 2x + 1} \right)^\prime = {e^x} + 2,$
$y'\left( {x = 0} \right) = f'\left( {x = 0} \right) = {e^0} + 2 = 3.$

By the inverse function theorem, we find:

$\varphi'\left( {y = 2} \right) = \frac{1}{{f'\left( {x = 0} \right)}} = \frac{1}{3}.$

### Example 13.

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

Solution.

Calculate the value of the original function and its derivative at the point $$x = 1:$$

$y\left( {x = 1} \right) = \sin 0 + {1^2} = 1,$
$y'\left( x \right) = f'\left( x \right) = \left[ {\sin \left( {x - 1} \right) + {x^2}} \right] = \cos \left( {x - 1} \right) + 2x,$
$y'\left( {x = 1} \right) = f'\left( {x = 1} \right) = \cos 0 + 2 \cdot 1 = 3.$

Now we can find the derivative of the inverse function $$x = \varphi \left( y \right):$$

$\varphi'\left( {y = 1} \right) = \frac{1}{{f'\left( {x = 1} \right)}} = \frac{1}{3}.$

### Example 14.

Find the derivative of the inverse function of $$y = {x^2} + 2\ln x$$ and calculate its value at $$x = 1.$$

Solution.

The original function $$y = f\left( x \right)$$ is defined for $$x \gt 0$$. Its derivative in this domain is positive:

$y' = f'\left( x \right) = \left( {{x^2} + 2\ln x} \right)^\prime = 2x + \frac{2}{x} \gt 0\;\;\text{for}\;\;x \gt 0.$

Hence, the function is monotonically increasing and has an inverse function. By the inverse function theorem, we have:

$\varphi'\left( y \right) = \frac{1}{{f'\left( x \right)}} = \frac{1}{{2x + \frac{2}{x}}} = \frac{1}{{\frac{{2{x^2} + 2}}{x}}} = \frac{x}{{2\left( {{x^2} + 1} \right)}}.$

In this case, the function $$x\left( y \right)$$ can not be expressed explicitly. However, using the last formula we can easily determine the value of the derivative of the inverse function at $$x = 1.$$ Preliminarily, we calculate the corresponding value of $$y:$$

$y\left( {x = 1} \right) = {1^2} + 2\ln 1 = 1 + 0 = 1.$

Then

$\varphi'\left( {y = 1} \right) = \frac{1}{{f'\left( {x = 1} \right)}} = \frac{1}{{2\left( {{1^2} + 1} \right)}} = \frac{1}{4}.$

### Example 15.

Find the derivative of the inverse function for $$y = {x^3} - 3x$$ and calculate its value at $$x = -2.$$

Solution.

Judging by the derivative:

$y' = f'\left( x \right) = \left( {{x^3} - 3x} \right)^\prime = 3{x^2} - 3 = 3\left( {{x^2} - 1} \right),$

the function has three intervals of monotonicity:

1. $$x \in \left( { - \infty , - 1} \right)$$ − the function is increasing;
2. $$x \in \left( { - 1, 1} \right)$$ − the function is decreasing;
3. $$x \in \left( {1, \infty} \right)$$ − the function is increasing.

Each interval can be mapped to its own inverse function. Further, we assume that the inverse function corresponds to the first interval containing the point $$x = -2.$$

The derivative of the inverse function has the form:

$\varphi'\left( y \right) = \frac{1}{{f'\left( x \right)}} = \frac{1}{{3\left( {{x^2} - 1} \right)}}.$

The function itself at $$x = -2$$ is equal

$y\left( {x = - 2} \right) = {\left( { - 2} \right)^3} - 3 \cdot \left( { - 2} \right) = - 8 + 6 = - 2.$

Then the value of the derivative of the inverse function at the indicated point is

$\varphi'\left( {y = - 2} \right) = \frac{1}{{f'\left( {x = - 2} \right)}} = \frac{1}{{3 \cdot \left( {{{\left( { - 2} \right)}^2} - 1} \right)}} = \frac{1}{9}.$

### Example 16.

Find the derivative of the inverse function for $$y = 3{x^2} - 1$$ and calculate its value at $$x = 2.$$

Solution.

Calculate the derivative of the given function:

$y' = f'\left( x \right) = \left( {3{x^2} - 1} \right)^\prime = 6x.$

As can be seen, the derivative changes sign when passing through the point $$x = 0$$, i.e. the function decreases for $$x \lt 0$$ and increases for $$x \gt 0$$. Further, we consider the branch containing the point $$x = 2$$. In this region, there exists an inverse function. Its derivative is given by the formula

$\varphi'\left( y \right) = \frac{1}{{f'\left( x \right)}}.$

Given that $$y\left( {x = 2} \right) = 3 \cdot {2^2} - 1 = 11,$$ we get:

$\varphi'\left( {y = 11} \right) = \frac{1}{{f'\left( {x = 2} \right)}} = \frac{1}{{6 \cdot 2}} = \frac{1}{{12}}.$

### Example 17.

$y = {\log _2}\left( {\frac{x}{3}} \right)$

Solution.

The function $$y = f\left( x \right) = {\log _2}\left( {\frac{x}{3}} \right)$$ is defined for $$x \gt 0$$ and monotonically increasing on this interval. Consequently, it has an inverse function $$x = \varphi \left( y \right):$$

$y = {\log _2}\left( {\frac{x}{3}} \right),\;\; \Rightarrow \frac{x}{3} = {2^y},\;\; \Rightarrow x = 3 \cdot {2^y}.$

By the inverse function theorem, we find:

$\left[ {{{\log }_2}\left( {\frac{x}{3}} \right)} \right]^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {3 \cdot {2^y}} \right)}^\prime }}} = \frac{1}{{3 \cdot {2^y} \cdot \ln 2}} = \frac{1}{{3\ln 2 \cdot {2^{{{\log }_2}\left( {\frac{x}{3}} \right)}}}} = \frac{1}{{3\ln 2 \cdot \frac{x}{3}}} = \frac{1}{{x\ln 2}}.$

Here we have used the logarithmic identity

${a^{{{\log }_a}x}} = x.$

### Example 18.

Find the derivative of $$y = \text{arcsec }x$$ at $$x = \sqrt 2.$$

Solution.

We use the fact that the arcsecant function is the inverse of the secant function. We assume that the derivative of secant is known:

$\left( {\sec y} \right)^\prime = \tan y\sec y = \frac{{\sin y}}{{{{\cos }^2}y}}.$

Take into account that the secant is equal to $$\sqrt 2$$ at $$y = {\frac{\pi }{4}}:$$

$\sec \left( {y = \frac{\pi }{4}} \right) = \frac{1}{{\cos \frac{\pi }{4}}} = \frac{1}{{\frac{{\sqrt 2 }}{2}}} = \sqrt 2.$

Then by the inverse function theorem, we have

$\left( {\text{arcsec }x} \right)^\prime = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\sec y} \right)}^\prime }}} = \frac{1}{{\tan y\sec y}} = \frac{{{{\cos }^2}y}}{{\sin y}}.$

Accordingly, the value of the derivative of $$\text{arcsec }x$$ at the point $$x = \sqrt 2$$ is equal:

$f'\left( {x = \sqrt 2 } \right) = \frac{1}{{\varphi'\left( {y = \frac{\pi }{4}} \right)}} = \frac{{{{\cos }^2}\frac{\pi }{4}}}{{\sin \frac{\pi }{4}}} = \frac{{{{\left( {\frac{{\sqrt 2 }}{2}} \right)}^2}}}{{\left( {\frac{{\sqrt 2 }}{2}} \right)}} = \frac{{\sqrt 2 }}{2}.$

### Example 19.

Find the derivative of the inverse function for $$y = x \cdot {3^x}$$ provided $$x \gt 0.$$

Solution.

Calculate the derivative of the given function:

$y' = f'\left( x \right) = \left( {x \cdot {3^x}} \right)^\prime = x' \cdot {3^x} + x \cdot {\left( {{3^x}} \right)^\prime } = 1 \cdot {3^x} + x \cdot {3^x}\ln 3 = {3^x}\left( {1 + x\ln 3} \right).$

It can be seen that the derivative is positive for $$x \gt 0$$. Hence, the function in this region monotonically increases and has an inverse function $$x = \varphi \left( y \right)$$. The derivative of the inverse function is expressed by the formula

$\varphi'\left( y \right) = \frac{1}{{f'\left( x \right)}} = \frac{1}{{{3^x}\left( {1 + x\ln 3} \right)}}.$

### Example 20.

Find the derivative of the inverse hyperbolic sine function $$y = \text{arcsinh } x.$$

Solution.

The functions $$y = \text{arcsinh } x$$ (inverse hyperbolic sine) and $$x = \sinh y$$ (hyperbolic sine) are mutually inverse functions. Therefore, by the inverse function theorem, we can write:

${\left( {\text{arcsinh }x} \right)^\prime } = f'\left( x \right) = \frac{1}{{\varphi'\left( y \right)}} = \frac{1}{{{{\left( {\sinh y} \right)}^\prime }}} = \frac{1}{{\cosh y}}.$

Express $$\cosh x$$ in terms of $$\sinh x$$ using the identity

${\cosh ^2}y - {\sinh ^2}y = 1$

(the hyperbolic analogue of the Pythagorean trigonometric identity).

From this, we find that

${\cosh ^2}y = 1 + {\sinh ^2}y,\;\; \Rightarrow \cosh y = \sqrt {1 + {{\sinh }^2}y}$

Given that $$\sinh \left( {\text{arcsinh }x} \right) = x,$$ we obtain the following expression for the derivative of the inverse hyperbolic sine function:

$\left( {\text{arcsinh }x} \right)^\prime = \frac{1}{{\cosh y}} = \frac{1}{{\sqrt {1 + {{\sinh }^2}y} }} = \frac{1}{{\sqrt {1 + {{\sinh }^2}\left( {\text{arcsinh }x} \right)} }} = \frac{1}{{\sqrt {1 + {x^2}} }}.$