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