Derivative Inverse Trigonometric Functions: Description, Formula table

Trigonometric function was discovered by Hipparchus, hence he’s called the father of trigonometry. Arcus functions, cyclometric functions, and anti-trigonometric functions are all called inverse trigonometric functions. The sin x, cos x, tan x, cot x, sec x, and cosec x are the six inverse trigonometric functions. These functions are used to calculate the angle of a trigonometric value. Inverse trigonometric functions are used in a variety of fields, including Physics, Mathematics, Engineering, Geometry, and Navigation. We'll study how to use implicit differentiation to obtain the derivative of inverse trigonometric functions in this article. But before we go any farther, let's cover implicit differentiation and inverse trigonometry.

Differentiation of inverse trigonometric functions

The inverse trigonometric function is represented by adding the power of -1 or by adding arc in prefix for a trigonometric function such as:

• Inverse of sin x = arcsin(x) or $Sin^{-1}x$

The inverse function theorem can be used to find the derivative inverse trigonometric functions. For example, consider the sine function $x=\varphi (y)=siny$ is the inverse trigonometric function for y = f(x) = arcsin x.

Derivative of inverse sin x = arcsin(x) or $Sin^{-1}x$ is,

$\Rightarrow \left ( arcsinx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( siny \right )^{,}}$

$\Rightarrow\frac{1}{cosy}$

$\Rightarrow \frac{1}{\sqrt{1-sin^{2}y}}$

$\Rightarrow \frac{1}{\sqrt{1-sin^{2}(arcsinx)}}$

$\Rightarrow \frac{1}{\sqrt{1-x^{2}}}$ ( -1 < x < 1)

Hence derivative of inverse of sin x is $ \frac{1}{\sqrt{1-x^{2}}}$ ( -1 < x < 1)

We may find the derivatives of the other inverse trigonometric functions using this method:

• For cos x function,

The derivative of cos inverse x is, $x=\varphi (y)=cosy$ is the inverse trigonometric function for y = f(x) = arc cos x.

Derivative of cos inverse [x = arc cos (x) ]or $cos^{-1}x$ is,

$\Rightarrow \left ( arccosx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( cosy \right )^{,}}$

$\Rightarrow\frac{1}{- siny}$

$ \Rightarrow - \frac{1}{\sqrt{1-cos^{2}y}}$

$ \Rightarrow - \frac{1}{\sqrt{1-cos^{2}(arccosx)}}$

$ \Rightarrow - \frac{1}{\sqrt{1-x^{2}}}$ ( -1 < x < 1)

Hence the cos inverse x differentiation is $ - \frac{1}{\sqrt{1-x^{2}}}$ ( -1 < x < 1)

• For tan x function,

The tan inverse derivative of x is, $x=\varphi (y)=tany$ is the inverse trigonometric function for y = f(x) = arc tan x.

Derivative of tan inverse x = arctan(x) or $tan^{-1}x$ is,

$\Rightarrow \left ( arctanx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( tany \right )^{,}}$

$\Rightarrow \frac{1}{sec^{2}y}$

$\Rightarrow \frac{1}{1+tan^{2}y}$

$\Rightarrow \frac{1}{1+tan^{2}(arctanx)}$

$\Rightarrow \frac{1}{1+x^{2}}$

Where $-\infty < x <\infty$

Hence the inverse derivative of tan x is $\Rightarrow \frac{1}{1+x^{2}}$

Where $-\infty < x <\infty$

• For cot x function,

The derivative of cot inverse x is, $x=\varphi (y)=coty$ is the inverse trigonometric function for y = f(x) = arc cot x.

Derivative of cot inverse [x = arccot(x)] or $cot^{-1}x$ is,

$\Rightarrow \left ( arccotx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( coty \right )^{,}}$

$\Rightarrow \frac{1}{- cosec^{2}y}$

$\Rightarrow \frac{1}{-(1+cot^{2}y)}$

$\Rightarrow \frac{1}{-(1+cot^{2}(arccotx))}$

$\Rightarrow \frac{1}{-(1+x^{2})}$

Where $-\infty < x <\infty$

Hence the inverse derivative of cot x is $\Rightarrow \frac{1}{-(1+x^{2})}$

Where $-\infty < x <\infty$

• For sec x function,

The differentiation of sec inverse x is, $x=\varphi (y)=secy$ is the inverse trigonometric function for y = f(x) = arc sec x.

Derivative of sec inverse [x = arc sec (x)] or $sec^{-1}x$ is,

$\Rightarrow \left ( arc secx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( secy \right )^{,}}$

$\Rightarrow\frac{1}{tany.secy}$

$\Rightarrow\frac{1}{secy\sqrt{sec^{2}y-1}}$

$\Rightarrow\frac{1}{|x|\sqrt{x^{2}-1}}$ where $x\in \left ( -\infty , -1 \right )\cup \left ( 1, \infty \right )$

In this formula, the absolute value x in the denominator emerges because the products tan y sec y should always be positive in the range of permissible value of y where $y\in \left ( 0 , \frac{\pi }{2} \right )\cup \left ( \frac{\pi }{2}, \pi \right )$, that is the derivative of the inverse secant (sec x) is always positive.

• Similarly, for cosine x function,

The derivative of inverse cosine of x is $x=\varphi (y)=cscy$ is the inverse trigonometric function for [y = f(x) = arccscx].

Inverse derivative of [cscx= arccsc(x)] or $csc^{-1}x$ is,

$\Rightarrow \left ( arccscx \right )^{,}=f^{,}(x)$

$\Rightarrow\frac{1}{\varphi (y)}$

$\Rightarrow\frac{1}{\left ( cscy \right )^{,}}$

$\Rightarrow- \frac{1}{ coty.cscy}$

$\Rightarrow - \frac{1}{cscy\sqrt{csc^{2}y-1}}$

$\Rightarrow - \frac{1}{|x|\sqrt{x^{2}-1}}$ where $x\in \left ( -\infty , -1 \right )\cup \left ( 1, \infty \right )$

Hence the inverse derivative of csc x is $ - \frac{1}{|x|\sqrt{x^{2}-1}}$ where $x\in \left ( -\infty , -1 \right )\cup \left ( 1, \infty \right )$

Table of differentiation of inverse trigonometric functions

Solved Examples

1. Differentiate: $T(p) = 2 cos p+ 6 cos^{-1}p$

Solution:

We know, $\frac{d}{dx}\left ( cos x \right )= - sinx$ and $\frac{d}{dx}\left ( cos^{-1}x \right )= - \frac{1}{\sqrt{1-x^{2}}}$

Therefore,

$T^{,}(p)=-2 sin (p)-\frac{6}{\sqrt{1-p^{2}}}$

2. Determine the derivative of $tan^{-1}\left ( sin^{-1}2x \right )$

Solution:

To answer this problem, we only need to employ the chain rule of differentiation and the equations for the derivatives of inverse trigonometric functions.

We know, $\frac{d}{dx}\left ( tan^{-1}x \right )=\frac{1}{1 + x^{2}}$ and $\frac{d}{dx}\left ( sin^{-1}x \right )= \frac{1}{\sqrt{1-x^{2}}}$

Therefore, $tan^{-1}\left ( sin^{-1}2x \right )= \frac{1}{\left\{1+ sin^{2}\left ( 2x \right ) \right\}^{2}}.\frac{1}{\sqrt{1-\left ( 2x \right )^{2}}}.2$

Conclusion

In geometric figures, trigonometric functions are used to calculate unknown angles and distances from known or measured angles. Trigonometry evolved as a result of the necessity to calculate angles and distances in professions such as astronomy, mapmaking, surveying, and artillery range finding. Trigonometry, being one of the essential disciplines of mathematics, is something that every student should study. Students with strong trigonometry abilities can calculate complicated angles and dimensions in a short amount of time.