Derivative of sin inverse x is 1/√(1-x²). The derivative of any function gives the rate of change of the functional value with respect to the input variable. Sin inverse x is one of the inverse trigonometric functions. It is also represented as sin^-1x. There are inverse trigonometric functions corresponding to each trigonometric function. The derivative of a function also helps in finding the slope of the tangent to the curve represented by the function at any point.

In this article, we will learn about the derivative of sin inverse x, methods to find it including the first principle of differentiation and implicit differentiation, solved examples, and practice problems.

What is the Derivative of Sin Inverse x?

The derivative of sin inverse x is 1/√(1-x²). It implies that the rate of change of the function, f(x) = sin^-1x with respect to the input variable, i.e. x, is 1/√(1-x²). Also, the slope of the curve represented by y = sin^-1x at any point x is given by dy/dx{Sin^-1x} = 1/√(1-x²). Thus, the formula for the derivative of sin inverse x can be written as follows:

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• Calculus in Maths
• Derivatives

Derivative of Sin Inverse x Formula

Formula for derivative of sin inverse x is given below:

(d/dx) [sin^-1x] = 1/√(1-x²)

or

(sin^-1x)’ = 1/√(1-x²)

It can be derived using the first principle of differentiation and implicit differentiation discussed as follows.

Proof of Derivative of Sin Inverse x

The derivative of sin inverse x can be found by two methods:

• Using First Principle of Differentiation
• Using Implicit Differentiation

Derivative of Sin Inverse x by First Principle of Differentiation

The first principle of differentiation states that derivative of a function f(x) is defined as,

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

or

f'(x) = lim_h→0 [f (x + h) – f(x)]/ h

Putting f(x) = sin^-1x to find derivative of sin inverse x, we get,

f'(x) = lim_h→0 [sin^-1(x + h) – sin^-1(x)]/ h

Putting, A = sin^-1(x + h) and B = sin^-1(x), we get, h = sin A – sin B and limit changes to A →B,

⇒ f'(x) = Lim _A→B (A – B) / (sin A – sin B)

Using trigonometric relation, sin A – sin B = 2(sin(A-B)/2) (cos(A+B)/2) we get,

⇒ f'(x) = Lim _A→B (A – B)/2(sin(A-B)/2) × 1/cos(A+B)/2

Using limit relation, lim_x→0 (sin x)/x = 1, we get,

⇒ f'(x) = Lim _A→B 1/cos(A+B)/2

⇒ f'(x) = 1/cos B

⇒ f'(x) = 1/√(1-sin²B)

⇒ f'(x) = 1/√(1-x²)

Hence, the formula for derivative of sin inverse x has been derived using first principle of differentiation.

Derivative of Sin Inverse x Implicit Differentiation

Implicit differentiation is used for the functions represented as y = f(x), where it is complex to find derivative of f(x) and it is relatively easier to find the derivative of g(y). Hence, the function is represented as x = g(y). This method is useful for calculating derivative of inverse functions and logarithmic functions. The derivative of sin inverse x is derived using this method as follows.

Let, y = sin^-1x

Then, sin y = x

Differentiating on both sides of above equation, we get,

⇒ cos y dy = dx

⇒ dy/dx = 1/cos y

Now, cos y = √(1-sin²y) = √(1-x²)

⇒ dy/dx = 1/√(1-x²)

Thus, we have derived the derivative of sin inverse x using implicit differentiation.

Read More,

• Derivative of Inverse Trigonometric Functions
• Differentiation of Trigonometric Functions
• Derivative Formulas
• Derivative of Arcsin x

Examples on Derivative of sin inverse x

Example 1: Find derivative of function represented as f(x) = sin^-1(2x).

Solution:

We know that, (sin^-1x)’ = 1/√(1-x²)

By chain rule, we get,

⇒ (sin^-12x)’ = 1/√(1-(2x)²) × d/dx(2x)

⇒ (sin^-12x)’ = 2/√(1-(2x)²)

⇒ (sin^-12x)’ = 2/√(1-4x²)

Thus, for f(x) = sin^-1(2x), we get, f'(x) = 2/√(1-4x²)

Example 2: Find derivative of function, f(x) = sin^-1 (x²).

Solution:

We have, f(x) = sin^-1 (x²)

We know that (sin^-1x)’ = 1/√(1-x²)

Thus, by applying chain rule, we get,

⇒ f'(x) = 1/√(1-x⁴) × d/dx(x²)

⇒ f'(x) = [1/√(1-x⁴)]×2x

⇒ f'(x) = 2x/√(1-x⁴)

Example 3: If p(x) = x²sin^-1x, find p'(x).

Solution:

For function p(x), we see that two functions are given in product. Thus, we use product rule of differentiation to obtain p'(x).

⇒ p'(x) = x² × d/dx(sin^-1x) + d/dx(x²) × sin^-1x

⇒ p'(x) = x²/√(1-x²) + 2xsin^-1x

Thus, for p(x) = x²sin^-1x, we have, p'(x) = x²/√(1-x²) + 2xsin^-1x

Example 4: Determine slope of tangent drawn to curve represented by y = sin^-1x at x = 1/√2.

Solution:

We know that slope to tangent at any curve is given dy/dx.

Thus, for y = sin^-1x, we have,

dy/dx = 1/√(1-x²)

Now, slope of tangent to curve at x = 1/√2 would be,

(dy/dx) _{x = 1/√2} = 1/√ (1-(1/√2)²)

(dy/dx) _{x = 1/√2} = √2

Thus, slope of tangent to curve y = sin^-1x at x=1/√2 would be √2.

Example 5: Find the derivative of function given by f(x) = sin^-1 (cos x).

Solution:

We know, (sin^-1x)’ = 1/√(1-x²) and (cos x)’ = -sin x

Therefore, by applying chain rule of differentiation for f(x) = sin^-1 (cos x), we get,

f'(x) = 1/√(1-cos²x)×d/dx(cos x)

f'(x) = -sin x/sin x = -1

Thus, f'(x) comes out to be -1.

Alternatively,

f(x) = sin^-1(cos x) = sin^-1(sin (π/2 – x))

f(x) = π/2 – x

Thus, f'(x) = -1

Practice Questions on Derivative of Sin Inverse x

Some practice questions on Derivative of Sin Inverse x

1. Find the derivative of the function f(x) = sin^-1x + cos^-1x

2. Find the derivative of the function f(x) = sin^-1 √x

3. Find the value of f'(x), if f(x) = xsin^-1x.

4. If y = sin^-1 (sin2x), then find the value of dy/dx.

5. If y = x/sin^-1x, find the value of dy/dx.

FAQs on Derivative of Sin inverse x

What is Derivative of a Function Imply?

Derivative of a function implies the change in the functional value with respect to the change in input variable. For physical quantities, derivative gives the rate of change of the quantity with input variables.

What is Derivative of Sin Inverse x?

Derivative of sin inverse x is 1/√(1-x²).

What are Different Methods to Find Derivative of Sin Inverse x?

Methods to find derivative of sin inverse x are as follows:

• First Principle of Differentiation
• Implicit Differentiation

What is the Application of Derivative of Sin Inverse x?

Derivative of sin inverse x is helpful in determining rate of change of inverse trigonometric functions with respect to the change in input variable.

What is the Derivative of Sin^-1(x²)?

Derivative of sin^-1(x²) is 2x/√(1-x⁴)