# Derivatives of Inverse Trigonometric Functions

Proofs of the formulas of the derivatives
of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Another method to find the derivative of inverse functions is also included and may be used.

## 1 - Derivative of \( y = \arcsin(x) \)

Let
\[ y = \arcsin(x) \]
which may be written as

\[ x = \sin(y) \]

Differentiate both side of the above with respect to \( x \)
\[ \dfrac{dx}{dx} = \dfrac{d (\sin(y))} {dx} \]
Simplify the left side and use the chain rule on the right hand side
\[ 1 = \cos(y) \dfrac{dy}{dx} \]
The above gives
\[ \dfrac{dy}{dx} = \dfrac{1}{\cos y} \]
The trigonometric identity \( \sin^2 y + \cos^2 y = 1\) gives
\[ \cos(y) = \sqrt{1 - \sin^2(y)} \]
From above, we have \( x = \sin(y) \) , hence
\[ \cos(y) = \sqrt{1 - x^2} \]

Substitute \( \cos(y) = \sqrt{1 - x^2} \) in \( \dfrac{dy}{dx} = \dfrac{1}{\cos y} \) to obtain
\[ \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} \]
Hence
\[
\Large \color{red}{\dfrac{d(\arcsin(x))}{dx} = \dfrac{1}{\sqrt{1 - x^2}}}
\]

## 2 - Derivative of \( \arccos(x) \)

Let
\[
y = \arccos(x)
\]
which may be written as
\[ x = \cos(y)\]
The differentiation of both sides of the above, with respect to \( x \) using the chain rule on the right hand side, gives
\[
1 = - \sin(y) \dfrac{dy}{dx}
\]
The above gives
\[
\dfrac{dy}{dx} = - \dfrac{1}{\sin(y) }
\]
The trigonometric identity \( \sin^2 y + \cos^2 y = 1\) gives
\[ \sin(y) = \sqrt{1 - \cos^2 (y)} \]
Use \( x = \cos(y)\) from above to write
\[
\sin(y) = \sqrt{1 - x^2}
\]
Substitute \( \sin y \) in \( \dfrac{dy}{dx} = - \dfrac{1}{\sin(y) } \) to obtain
\[
\Large \color{red}{\dfrac{d(\arccos(x))}{dx} = - \dfrac{1}{\sqrt{1 - x^2}}}
\]

## 3 - Derivative of \( \arctan(x) \)

Let
\[
y = \arctan(x)
\]
which may be written as
\[
x = \tan(y)
\]
We differentiate both side with respect to x, using the chain rule on the right hand side, to obtain
\[
1 = \sec^2(y) \dfrac{dy}{dx}
\]
The above gives
\[
\dfrac{dy}{dx} = \dfrac{1}{\sec^2(y) } = \cos^2(y)
\]
We now use the trignometric identity
\[
\cos^2(y) = \dfrac{1}{1+\tan^2(y)}
\]
and \( x = \tan(y) \) from above to express \( \sec^2(y) \) in terms of x as follows
\[
\cos^2(y) = \dfrac{1}{1+x^2}
\]
Substitute in Hence \( \dfrac{dy}{dx} = \cos^2(y) \) to obtain
\[
\Large \color{red}{\dfrac{d(\arctan(x))}{dx} = \dfrac{1}{1+x^2 }}
\]

## 4 - Derivative of \( \text{arccot}(x) \)

Let
\[
y = \text{arccot}(x)
\]
which may be written as
\[
x = \cot(y)
\]
Differentiate both side with respect to x, using the chain rule on the right hand side, gives
\[
1 = - \csc^2(y) \dfrac{dy}{dx}
\]
The above gives
\[
\dfrac{dy}{dx} = - \dfrac{1}{\csc^2(y) } = - \sin^2(y)
\]
Use the trigonometric identity
\[
\sin^2(y) = \dfrac{1}{1 + \cot^2(y)}
\]
and \( x = \cot(y) \) from above to express \( \sin^2(y) \) in terms of x as follows
\[
\sin^2(y) = \dfrac{1}{1 + \cot^2(y)} = \dfrac{1}{1 + x^2}
\]
Hence
\[
\Large \color{red}{\dfrac{d(\text{arccot}(x))}{dx} = - \dfrac{1}{1+x^2 }}
\]

## 5 - Derivative of \( \text{arcsec}(x) \)

Let
\[
y = \text{arcsec}(x)
\]
which may be written as
\[
x = \sec(y)
\]
The differentiation of both side with respect to x, using the chain rule on the right hand side, gives
\[
1 = \sec(y) \tan(y) \dfrac{dy}{dx}
\]
The above gives
\[
\dfrac{dy}{dx} = \dfrac{1}{\sec(y) \tan(y) }
\]
Use the identity \( \tan y = \sqrt{\sec^2 y - 1} \) and \( x = \sec(y) \) from above to
express \( \sec(y) \tan(y) \) in terms of x as follows
\[
\sec(y) \tan(y) = x \sqrt{x^2 - 1}
\]
Hence
\[
\Large \color{red}{\dfrac{d(\text{arcsec}(x))}{dx} = \dfrac{1}{x \sqrt{x^2 - 1} }}
\]

## 6 - Derivative of \( \text{arccsc}(x) \)

Let
\[
y = \text{arccsc}(x)
\]
which may be written as
\[
x = \csc(y)
\]
The differentiation of the left and right sides of the above, using the chain rule on the right hand side, gives
\[
1 = - \csc(y) \cot(y) \dfrac{dy}{dx}
\]
The above gives
\[
\dfrac{dy}{dx} = - \dfrac{1}{ \csc(y) \cot(y) }
\]
Use the trigonometric identity \( \cot y = \sqrt {csc^2 y - 1} \) and \( x = \csc(y) \) from above
to express \( \csc(y) \cot(y) \) in terms of x as follows
\[
\csc(y) \cot(y) = \csc(y) \sqrt {\csc^2 y - 1} = x \sqrt{x^2 - 1}
\]
Hence
\[
\Large \color{red}{\dfrac{d(\text{arccsc}(x))}{dx} = - \dfrac{1}{x \sqrt{x^2 - 1} }}
\]

## Examples with Solutions

### Example 1

Find the first 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 \( g \): \( f(x) = h(x) g(x) \).

Use the product rule of differentiation: \( f '(x) = h(x) g '(x) + g(x) h '(x) \), to differentiate function \( f \) as follows

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

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

### 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 \[ f(x) = \arcsin (2x + 2) \]

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

\( g '(x) = 2 \left( \dfrac{1}{\sqrt{1 - u^2}} \right) \)

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

## More References and links

Derivative of Inverse Function

differentiation and derivatives