Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the trigonometric functions. They are used to determine the angle that corresponds to a given trigonometric ratio. Differentiation of these functions is a common topic in calculus, and understanding their derivatives is essential for solving various problems in mathematics, physics, and engineering.

Inverse Trigonometric Functions and Their Derivatives

The main inverse trigonometric functions are:

1. Arcsine function ($\arcsin x$ or $\sin^{-1} x$)
2. Arccosine function ($\arccos x$ or $\cos^{-1} x$)
3. Arctangent function ($\arctan x$ or $\tan^{-1} x$)
4. Arccotangent function ($\text{arccot} x$ or $\cot^{-1} x$)
5. Arcsecant function ($\text{arcsec} x$ or $\sec^{-1} x$)
6. Arccosecant function ($\text{arccsc} x$ or $\csc^{-1} x$)

The derivatives of these functions are derived from the derivatives of their respective trigonometric functions using the chain rule and implicit differentiation.

Table of Derivatives

Function	Derivative	Domain of Derivative
$\arcsin x$	$\frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arccos x$	$\frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1-x^2}}$	$-1 < x < 1$
$\arctan x$	$\frac{d}{dx}(\arctan x) = \frac{1}{1+x^2}$	$-\infty < x < \infty$
$\text{arccot} x$	$\frac{d}{dx}(\text{arccot} x) = -\frac{1}{1+x^2}$	$-\infty < x < \infty$
$\text{arcsec} x$	$\frac{d}{dx}(\text{arcsec} x) = \frac{1}{	x
$\text{arccsc} x$	$\frac{d}{dx}(\text{arccsc} x) = -\frac{1}{	x

Formulas and Derivations

The derivatives of the inverse trigonometric functions can be derived using implicit differentiation. For example, let's derive the derivative of the arcsine function:

1. Start with the definition of the arcsine function: $y = \arcsin x$.
2. Take the sine of both sides: $\sin y = x$.
3. Differentiate implicitly with respect to $x$: $\cos y \frac{dy}{dx} = 1$.
4. Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{1}{\cos y}$.
5. Since $\sin^2 y + \cos^2 y = 1$, we have $\cos y = \sqrt{1 - \sin^2 y}$.
6. Substitute $\sin y = x$ into the equation: $\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}$.

This process can be repeated for each of the inverse trigonometric functions to find their derivatives.

Examples

Example 1: Derivative of $\arcsin x$

Find the derivative of $f(x) = \arcsin(2x)$.

Solution:

Using the chain rule and the derivative of $\arcsin x$, we get:

$$ f'(x) = \frac{d}{dx}(\arcsin(2x)) = \frac{1}{\sqrt{1-(2x)^2}} \cdot 2 = \frac{2}{\sqrt{1-4x^2}} $$

Example 2: Derivative of $\arctan x$

Find the derivative of $f(x) = \arctan(\sqrt{x})$.

Solution:

Using the chain rule and the derivative of $\arctan x$, we get:

$$ f'(x) = \frac{d}{dx}(\arctan(\sqrt{x})) = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2x + 2\sqrt{x}} $$

Example 3: Derivative of $\text{arcsec} x$

Find the derivative of $f(x) = \text{arcsec}(3x)$.

Solution:

Using the chain rule and the derivative of $\text{arcsec} x$, we get:

$$ f'(x) = \frac{d}{dx}(\text{arcsec}(3x)) = \frac{1}{|3x|\sqrt{(3x)^2-1}} \cdot 3 = \frac{3}{|3x|\sqrt{9x^2-1}} $$

Note that the absolute value in the denominator is necessary because the domain of the arcsecant function excludes the interval $(-1, 1)$.

Understanding the derivatives of inverse trigonometric functions is crucial for solving problems involving integration, curve sketching, and many other applications in calculus.