Derivative of inverse sine | Taking derivatives | Differential Calculus | Khan Academy
Derivative of inverse sine | Taking derivatives | Differential Calculus | Khan Academy

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Derivative of Arcsin

To find the derivative of arcsin, we have to consider some facts about arcsin. arcsin (which can also be written as sin-1) is the inverse function of the sine function. i.e., If y = sin-1x then sin y = x. By the definition of inverse functions, if f and f-1 are inverse functions of each other then f(f-1(x)) = f-1(f(x)) = x. From this, sin(arcsin x) = arcsin(sin x) = x in the corresponding domains. We use these facts to find the derivative of arcsin x.

Let us see the formula of derivative of arcsin along with proof (in two different methods) and few solved examples.

 1 What Is Derivative of arcsin? 2 Derivative of arcsin Proof by Chain Rule 3 Derivative of arcsin Proof by First Principle 4 FAQs on Derivative of arcsin

## What Is Derivative of arcsin?

The derivative of arcsin x is denoted by d/dx(arcsin x) (or) d/dx(sin-1x) (or) (arcsin x)’ (or) (sin-1x)’. Its value is 1/√1 – x². We will prove it in two methods in the upcoming sections. The two methods are

• Using the chain rule
• Using the first principles

### Derivative of arcsin x Formula

The derivative of the arcsin function is,

• d/dx(arcsin x) = 1/√1 – x² (OR)
• d/dx(sin-1x) = 1/√1 – x²

We will prove this formula now in the next sections in each of the above-mentioned methods.

## Derivative of arcsin Proof by Chain Rule

To find the derivative of arcsin using the chain rule, assume that y = arcsin x. Taking sin on both sides,

sin y = sin (arcsin x)

By the definition of inverse function, sin (arcsin x) = x. So the above equation becomes,

sin y = x … (1)

Differentiating both sides with respect to x,

d/dx (sin y) = d/dx(x)

We have d/dx (sin x) = cos x. Also, by chain rule,

cos y · dy/dx = 1

dy/dx = 1/cos y

Using one of the trigonometric identities, cos y = √1 – sin²y = √1 – x² (from (1))

So dy/dx = 1/√1 – x²

Putting y = arcsin x back here,

d/dx (arcsin x) = 1/√1 – x²

Hence proved.

## Derivative of arcsin Proof by First Principle

Let us recall that the derivative of a function f(x) by the first principle (definition of the derivative) is given by the limit, f'(x) = limₕ→₀ [f(x + h) – f(x)] / h. To find the derivative of arcsin x, assume that f(x) = arcsin x. Then f(x + h) = arcsin (x + h). Then from the above limit,

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

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

Then sin A = x + h and sin B = x.

Subtracting the second equation from the first,

sin A – sin B = (x + h) – x

sin A – sin B = h

Since h → 0, (sin A – sin B) → 0. From this, sin A → sin B (or) A → B (or) A – B → 0.

Then the above limit becomes,

f'(x) = lim$$_{A-B → 0}$$ (A – B) / (sin A – sin B)

We have sin A – sin B = 2 sin [(A – B)/2] cos [(A + B)/2]. Substituting this in the above limit,

f'(x) = lim$$_{A-B→ 0}$$ (A – B) / [2 sin [(A – B)/2] cos [(A + B)/2] ]

When A – B → 0, we can have (A – B)/2 → 0.

f'(x) = lim$$_{\frac{A-B}{2}→ 0}$$ [(A – B)/2] / sin [(A – B)/2] lim$$_{A → B}$$ cos[(A + B)/2]

f'(x) = lim$$_{\frac{A-B}{2}→ 0}$$ 1/ (sin [(A – B)/2] / [(A – B)/2] ) lim$$_{A → B}$$ cos[(A + B)/2]

We know that lim ₓ→₀ (sin x)/x = 1. From this,

f'(x) = (1/1) cos[(B + B)/2] = cos B … (2)

We have sin B = x.

So cos B = √1 – sin²B = √1 – x²

Therefore, f'(x) = 1/√1 – x²

Hence proved.

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Here are some topics that are related to the derivative of arcsin x.

## Solved Examples Using Derivative of Arcsin x

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

Solution:

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

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

Also, by chain rule,

y’ = 1/√(1 – (1/x)² · d/dx (1/x)

= 1/ √(1 -(1/x²) · (-1/x2)

= x/√x² – 1 · (-1/x2)

= (-1) / (x√x² – 1)

Answer: The derivative of the given function is (-1) / (x√x² – 1).

2. Example 2: Find the derivative of y = sin-1(1 + x2).

Solution:

Let f(x) = sin-1(1 + x2).

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

Also, by chain rule,

y’ = 1/√1 – (1+ x²)² · d/dx (1 + x2)

= 1 / √1 – 1 – x⁴ – 2x² · (2x)

= (2x) / √- x⁴ – 2x²

= (2x) / √x²(- x² – 2)

= 2 / √-x² -2

Answer: The derivative of the given function is 2 / √- x² – 2.

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

Solution:

The given function is, y = x arcsin x.

It is a product. So we use the product rule to find its derivative.

y’ = x · d/dx (arcsin x) + arcsin x · d/dx (x)

= x [1/√1-x²] + arcsin x (1)

= x/√1-x² + arcsin x

Answer: The derivative of the given function is x/√1-x² + arcsin x.

## FAQs on Derivative of Arcsin

### What is Derivative of arcsin?

The derivative of arcsin x is 1/√1-x². It is written as d/dx(arcsin x) = 1/√1-x². This also can be written as d/dx(sin-1x) = 1/√1-x².

### How to Find the Derivative of arcsin?

To derive the derivative of arcsin, assume that y = arcsin x then sin y = x. Differentiating both sides with respect to y, then cos y = dx/dy. Taking reciprocals, dy/dx = 1/(cos y) = 1/√1 – sin²y = 1/√1-x².

### What is the Derivative of arcsin √x?

We know that the derivative of arcsin x to be 1/√1 – x². By using this formula and chain rule, the d/dx(arcsin √x) = 1/√1 – (√x)² d/dx (√x) = 1/√1 – x · (1/2√x) = 1/(2√ x- x²)).

### What is the Derivative of arcsin x/2?

We have the derivative of arcsin x to be 1/√1 – x². By using this and chain rule, the d/dx(arcsin x/2) = √1 – (x/2)² d/dx (x/2) = √1 – x²/4 · (1/2) = 2/√4 – x² · (1/2) = 1/√4 – x².

### Is arcsin the Derivative of Sin?

No, the derivative of arcsin is NOT sin. The derivative of arcsin x is 1/√1 – x².

### What is the Derivative of arcsin x/a?

The derivative of arcsin x to be 1/√1 – x². By using this and chain rule, the d/dx(arcsin x/a) = 1/√1 – (x/a)² d/dx (x/a) = 1/√1 – (x²/a²) · (1/a) = [a/√a² – x²] · (1/a) = 1/√a² – x².

What is the Difference Between the Derivatives of sin x and arcsin x?

The derivative of sin x is cos x whereas the derivative of arcsin x is 1/√1-x².

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