Differentiation of Inverse Trigonometric Functions

In this article, we will explore the application of implicit differentiation to find the derivative of inverse trigonometric functions. But before heading forward, let’s brush up on the concept of implicit differentiation and inverse trigonometry.

Inverse Trigonometry

Inverse trigonometric functions are the inverse functions of the trigonometric ratios i.e. sin, cos, tan, cot, sec, cosec. These functions are widely used in fields like physics, mathematics, engineering, and other research fields. Just like addition and subtraction are the inverses of each other, the same is true for the inverse of trigonometric functions.

sin θ = x
⇒ θ = sin⁻¹x

Representation of Inverse Trigonometric Functions

They are represented by adding arc in prefix or by adding -1 to the power.

Inverse sine can be written in two ways:
sin^-1x
arcsin x
Same goes for cos and tan.

Note: Don’t confuse sin^-1x with (sin x)^-1. They are different. Writing sin^-1 x is a way to write inverse sine whereas (sin x)^-1 means 1/sin x.

Implicit Differentiation

Implicit differentiation is a method that makes use of the chain rule to differentiate implicitly defined functions. It is generally not easy to find the function explicitly and then differentiate. Instead, we can totally differentiate f(x, y) and then solve the rest of the equation to find the value of f'(x). Even when it is possible to explicitly solve the original equation, the formula resulting from total differentiation is, in general, much simpler and easier to use. Let’s differentiate some of the inverse trigonometric functions.

Example 1: Differentiate sin^-1(x)?

Solution:

Let,
y = sin⁻¹(x)
Taking sine on both sides of equation gives,
$sin (y) = sin (sin^{-1}(x))$
By the property of inverse trigonometry we know, $sin (sin^{-1}(x))=x$
$sin (y) = x$
Now differentiating both sides wrt to x,
$\frac{d}{dx}sin (y)= \frac{d}{dx}x$
$cos(y). \frac{dy}{dx} = 1$
$\frac{dy}{dx} = \frac{1} { cos(y)}$
We can simplify it more by using the below observation:
$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\x^2+cos^2(y)&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ sin (y) = x )\\cos^2(y)&=1-x^2\\ cos(y)&=\sqrt{1-x^2}\\ \end{aligned}$
Substituting the value, we get
$\begin{aligned} \frac{dy}{dx}&=\frac{1}{cos(y)} \\ \frac{dy}{dx}&=\frac{1}{\sqrt{1-x^2}} \end{aligned}$

Example 2: Differentiate cos^-1(x)?

Solution:

Let,
y = cos⁻¹(x)
Taking cosine on both sides of equation gives,
$cos (y) = cos (cos^{-1}(x))$
By the property of inverse trigonometry we know, $cos (cos^{-1}(x))=x$
$cos (y) = x$
Now differentiating both sides wrt to x,
$\frac{d}{dx}cos (y)= \frac{d}{dx}x$
$-sin(y). \frac{dy}{dx} = 1$
$\frac{dy}{dx} = \frac{-1} { sin(y)}$
We can simplify it more by using the below observation:
$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\sin^2(y)+x^2&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ cos (y) = x )\\sin^2(y)&=1-x^2\\ sin(y)&=\sqrt{1-x^2}\\ \end{aligned}$
Substituting the value, we get
$\begin{aligned} \frac{dy}{dx}&=\frac{-1}{sin(y)} \\ \frac{dy}{dx}&=\frac{-1}{\sqrt{1-x^2}} \end{aligned}$

Example 3: Differentiate tan^-1(x)?

Solution:

Let,
y = tan⁻¹(x)
Taking tan on both sides of equation gives,
$tan (y) = tan (tan^{-1}(x))$
By the property of inverse trigonometry we know, $tan (tan^{-1}(x))=x$
$tan (y) = x$
Now differentiating both sides wrt to x,
$\frac{d}{dx}tan (y)= \frac{d}{dx}x$
$sec^2(x). \frac{dy}{dx} = 1$
$\frac{dy}{dx} = \frac{1} {sec^2(x)}$
We can simplify it more by using the below observation:
$\begin{aligned} sec^2(y)-tan^2(y)&=1 \\sec^2(y)-x^2&=1 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ tan (y) = x )\\sec^2(y)&=1+x^2\end{aligned}$
Substituting the value, we get
$\begin{aligned} \frac{dy}{dx}&=\frac{1}{sec^2(y)} \\ \frac{dy}{dx}&=\frac{1}{{1+x^2}} \end{aligned}$

Some Advanced Examples of Inverse Trigonometry Functions Differentiation

Example 1: y = cos^-1 (-2x²). Find dy/dx at x = 1/2?

Solution:

Method 1 (Using implicit differentiation)

Given,
y = cos⁻¹(−2x²)
⇒ cos y = −2x²
Differentiating both sides wrt x
$\begin{aligned} \frac{d}{dx} cos(y) &=\frac{d}{dx} -2x^2 \\ -sin(y) \frac{dy}{dx} &=-4x \\ \frac{dy}{dx}&=\frac{4x}{sin (y)} \end{aligned}$
Simplifying
$\begin{aligned} sin^2(y)+cos^2(y)&=1 \\sin^2(y)+(-2x^2)^2 &=1 \\sin^2(y)+4x^4 &=1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( As\ cos (y) = -2x^2 )\\sin^2(y)&=1-4x^4\\ sin(y)&=\sqrt{1-4x^4}\\ \end{aligned}$
Putting the obtained value we get,
$\begin{aligned} \frac{dy}{dx} &=\frac{4x}{\sqrt{1-4x^4}} \\ &=\frac{4(\frac{1}{2})}{\sqrt{1-4(\frac{1}{2})^4}} \\ &=\frac{2}{\sqrt{1-\frac{1}{4}}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}$

Method 2 (Using chain rule as we know the differentiation of arccos x)

Given,
y = cos⁻¹(−2x²)
Differentiating both sides wrt x
$\begin{aligned} \frac{dy}{dx} &=\frac{d}{dx} cos^{-1}(-2x^2) \\ &=\frac{-1}{\sqrt{1-(-2x^2)^2}}\ .\ (-4x) \\ &=\frac{4x}{\sqrt{1-4x^4}} \\ &=\frac{4(\frac{1}{2})}{\sqrt{1-4(\frac{1}{2})^4}} \\ &=\frac{2}{\sqrt{1-\frac{1}{4}}} \\ &=\frac{4}{\sqrt{3}} \end{aligned}$

Example 2: Differentiate $\begin{aligned}sin^{-1}(\frac{1-x}{1+x}) \end{aligned}$ ?

Let,
$\begin{aligned} y = sin^{-1}(\frac{1-x}{1+x}) \end{aligned}$
Differentiating both sides wrt x
$\begin{aligned} \frac{dy}{dx} &= \frac{d}{dx}sin^{-1}(\frac{1-x}{1+x}) \\ &= \frac{1}{\sqrt{1-(\frac{1-x}{1+x})^2}} \ . \frac{d}{dx}(\frac{1-x}{1+x}) \\ &= \frac{1+x}{\sqrt{(1+x)^2-({1-x})^2}} \ . \frac{-(1+x)-(1-x)}{(1+x)^2} \\ &= \frac{1}{\sqrt{(1+x)^2-({1-x})^2}} \ . \frac{-2}{(1+x)} \\ &= \frac{1}{\sqrt{4x}} \ . \frac{-2}{(1+x)} \\ &= \frac{-1}{\sqrt{x}(1+x)} \\ \end{aligned}$

Table of Derivatives of Inverse Trigonometric Functions

Function	Derivative
$sin^{-1} x$	$\frac{1}{\sqrt{1-x^2}}$
$cos^{-1} x$	$\frac{-1}{\sqrt{1-x^2}}$
$tan^{-1} x$	$\frac{1}{{1+x^2}}$
$cot^{-1} x$	$\frac{-1}{{1+x^2}}$
$sec^{-1} x$	$\frac{1}{\|x\|\sqrt{x^2-1}}$
$cosec^{-1} x$	$\frac{-1}{\|x\|\sqrt{x^2-1}}$