Derivative of Arcsin (original) (raw)

Last Updated : 23 Jul, 2025

Derivative of Arcsin x is d/dx(arcsin x) = 1/√1-x². It is denoted by d/dx(arcsin x) or d/dx(sin-1x). Derivative of Arcsin refers to the process of finding the rate of change in Arcsin x function with respect to the independent variable. Derivative of Arcsin x is also known as differentiation of Arcsin.

In this article, we will learn about the derivative of Arcsin and its formula including the proof of the formula using the first principle of derivatives, quotient rule, and chain rule method.

Table of Content

What is Derivative in Math?

**Derivative of a function is the rate of change of the function with respect to any independent variable. The derivative of a function f(x) is denoted as f'(x) or (d /dx)[f(x)]. The differentiation of a trigonometric function is called a derivative of the trigonometric function or trig derivatives. The derivative of a function f(x) is defined as:

**f'(x 0 ) = lim h→0 [f(x 0 + h) – f(x 0 )] / h

**Read More:

What is Derivative of sin-1x?

Among the inverse trig derivatives, the derivative of the Arcsin x is one of the derivatives. Derivative of arcsin function represents the rate at which the arcsin curve is changing at a given point. It is denoted by d/dx(arcsin x) or d/dx(sin-1x). Arcsinx is also known as inverse sin x.

**Derivative of the Arcsin x is 1/√1-x²

Derivative of Arcsin x Formula

The formula for the derivative of Arcsin x is given by:

****(d/dx) [Arcsin x] = 1/√1-x²**

**OR

****(Arcsin x)’ = 1/√1-x²**

**Also Check, InverseTrigonometric Function

Proof of Derivative of Arcsin x

The derivative of tan x can be proved using the following ways:

**Derivative of Arcsin by Chain Rule

To prove derivative of Arcsin x by chain rule, we will use basic trigonometric and inverse trigonometric formula:

**Here is the proof of derivative of Arcsin x:

Let y = arcsinx

Taking sin on both sides

siny = sin(arcsinx)

By the definition of an inverse function, we have,

sin(arcsinx) = x

So the equation becomes siny = x .....(1)

Differentiating both sides with respect to x,

d/dx (siny) = d/dx (x)

cosy · d/dx(y) = 1 [ As d/dx(sin x) = cos x]

dy/dx = 1/cosy

Using one of the trigonometric identities

sin2 y+cos2 y = 1

∴ cos y = √1 – sin2y = √1–x2 [From (1) we have siny = x]

dy/dx = 1/√(1–x2)

Substituting y = arcsin x

**d/dx (arcsinx) = arcsin′x = 1/√1 - x 2

**Also Check, Chain Rule

Derivative of Arcsin by First Principle

To prove derivative of arcsin x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:

We can prove the derivative of arcsin by First Principle using the following steps:

Let f(x) = arcsinx

By First principle we have

\frac{d f( x)}{dx} =\displaystyle \lim_{h \to 0} \frac{f (x + h)- f(x)}{h}

put f(x) = arcsinx, we get

\frac{d}{dx}(arcsin x) =\displaystyle \lim_{h \to 0} \frac{arcsin (x + h)- arcsin x}{h}....(1)

Assume that arcsin (x + h) = A and arcsin x = B

So we have,

sin A = x+h .....(2)

sin B = x .......(3)

Subtract (3) from (2), we have

sin A – sinB = (x+h) – x

sinA – sinB = h

If h → 0, (sin A – sin B) → 0

sin A → sin B or A → B

Substitute these values in eq(1)

\frac{d}{dx}(arcsin x) =\displaystyle \lim_{A \to B} \frac{A- B}{Sin A- Sin B}

Using sin A – sin B = 2 sin [(A – B)/2] cos [(A + B)/2], we get

\frac{d}{dx}(arcsin x) =\displaystyle \lim_{A \to B} \frac{A- B}{2Cos \frac{A+B}{2}- 2 Sin \frac{A-B}{2}}

which can be written as:

\frac{d}{dx}(arcsin x) =\displaystyle \lim_{A \to B} \frac{\frac{A- B}{2}}{Sin \frac{A-B}{2}}\times \frac{1}{Cos \frac{A+B}{2}}

Now, we know limx→0 x/sinx = 1, therefore the above equation changes to

\frac{d}{dx}(arcsin x) ={1}\times \frac{1}{Cos \frac{B+B}{2}}

\frac{d}{dx}(arcsin x) =\frac{1}{Cos {B}}

Using one of the trigonometric identities

sin2 y+cos2 y = 1

∴ cos B = √1 – sin2B = √1–x2 [Sin B = x from (3)]

f′(x) = dy/dx = 1 / √(1–x2)

**Also, Check

Solved Examples on Derivative of Arcsin x

**Example 1: Find the derivative of y = arcsin (3x).

**Solution:

Let f(x) = arcsin (3x).

We know that d/dx (arcsin x) = 1/√1 - x².

By chain rule,

d/dx(arcsin(3x)) = 1/√(1 - (3x)² · d/dx (3x)

= 1/ √(1 -9x²) · (3)

= 3/√(1 -9x²)

Hence, the derivative of y = arcsin (3x) is 3/√(1 -9x²).

**Example 2: Find the derivative of y = arcsin (1/2x).

**Solution:

Let f(x) = arcsin (1/2x).

We know that d/dx (arcsin x) = 1/√1 - x².

By chain rule,

d/dx(arcsin(1/2x)) = 1/√(1 - (1/2x)² · d/dx (1/2x)

= 1/ √(1 -(1/4x²) )· (-1/2x2)

= 1/√(4x2 - 1)/4x2 · (-1/2x2)

= -1/x√4x2 - 1

Hence, the derivative of y = arcsin (1/x) is -1/x√4x2 - 1.

**Example 3: Find the derivative of y = x arcsin x.

**Solution:

We have y = x arcsin x.

d/dx(arcsin(1/x)) = x · d/dx (arcsin x) + arcsin x · d/dx (x)

= x [1/√1-x²] + arcsin x (1)

= x/√1-x² + arcsin x
Hence, the derivative of y = arcsin (1/x) is x/√1-x² + arcsin x

Practice Questions on Derivative of Sin x

**1. Find the derivative of sin -1 (5x).

**2. Find the derivative of x 3 sin -1 (x).

**3. Evaluate: d/dx [ sin -1 (x) / x 2 **+ 1 ]

**4. Evaluate the derivative of sin -1 (x) - tan(x)

Summary

To find the derivative of \arcsin(x), we start by letting y = \arcsin(x). This implies that \sin(y) = x. Differentiating both sides with respect to x using implicit differentiation, we get \sin(y) = x\sin(y) = x. Solving for \frac{dy}{dx}​, we have \frac{dy}{dx} = \frac{1}{\cos(y)}​. Since \cos(y) = \sqrt{1 - \sin^2(y)}​ and \sin(y) = x, it follows that \cos(y) = \sqrt{1 - x^2}​. Thus, the derivative of \arcsin(x) is \frac{d}{dx} (\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}​.