# Derivatives of Inverse Trigonometric Functions

Formulas of the derivatives, in calculus, of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions.

## 1 - Derivative of $\arcsin x$.

The derivative of $f(x) = \arcsin x$ is given by

$f \, '(x) = \dfrac{1}{\sqrt{1-x^2}}$

## 2 - Derivative of $\arccos x$.

The derivative of $f(x) = \arccos x$ is given by

$f \, '(x) = - \dfrac{1}{\sqrt{1-x^2}}$

## 3 - Derivative of $\arctan x$.

The derivative of $f(x) = \arctan x$ is given by

$f \, '(x) = \dfrac{1}{1+x^2}$

## 4 - Derivative of $\text{arccot} \, x$.

The derivative of $f(x) = \text{arccot} \, x$ is given by

$f \, '(x) = - \dfrac{1}{1 + x^2}$

## 5 - Derivative of $\text{arcsec}\, x$.

The derivative of $f(x) = \text{arcsec}\, x$ is given by

$f \, '(x) = \dfrac{1}{|x| \sqrt{x^2 - 1}}$

## 6 - Derivative of $\text{arccsc}\, x$.

The derivative of $f(x) = \text{arccsc}\, x$ is given by

$f \, '(x) = - \dfrac{1}{|x| \sqrt{x^2-1}}$

Example 1: Find the derivative of $f(x) = x \arcsin x$

Solution to Example 1:

• Let $h(x) = x$ and $g(x) = \arcsin x$, function $f$ is considered as the product of functions $h$ and $g$: $f(x) = h(x) \cdot g(x)$. Hence we use the product rule, $f \, '(x) = h(x) g \, '(x) + g(x) h \, '(x)$, to differentiate function $f$ as follows

$f \, '(x) = x \dfrac{1}{\sqrt{1-x^2}} + \arcsin x \cdot 1 = \dfrac {x}{\sqrt{1-x^2}} + \arcsin x$

Example 2: Find the first derivative of $f(x) = \arctan x + x^2$

Solution to Example 2:

• Let $g(x) = \arctan x$ and $h(x) = x^2$, function $f$ may be considered as the sum of functions $g$ and $h$: $f(x) = g(x) + h(x)$. Hence we use the sum rule, $f \, '(x) = g \, '(x) + h \, '(x)$, to differentiate function $f$ as follows

$f \, '(x) = \dfrac{1}{1+x^2} + 2x = \dfrac{2x^3 + 2x + 1}{1+x^2}$

Example 3: Find the first derivative of $f(x) = \arcsin (2x + 2)$

Solution to Example 3:

• Let $u(x) = 2x + 2$, function $f$ may be considered as the composition $f(x) = \arcsin(u(x))$. Hence we use the chain rule, $f \, '(x) = \dfrac{du}{dx} \dfrac{d(arcsin(u))}{du}$, to differentiate function $f$ as follows

$f \, '(x) = 2 \cdot \dfrac{1}{\sqrt{1-u^2}}$

$= \dfrac{2}{\sqrt{1-(2x+2)^2}}$

More on differentiation and derivatives

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Updated: 2 April 2013