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# Calculus

Differentiation of Functions

# Derivatives of Inverse Functions

Page 1
Problems 1-3
Page 2
Problems 4-20

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,
$\varphi’\left( {{y_0}} \right) = \frac{1}{{f’\left( {{x_0}} \right)}},$ i.e. 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 on problem description to see 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
${e^{\ln x}} = x.$

Page 1
Problems 1-3
Page 2
Problems 4-20