The derivative of arcsin(x), also known as the inverse sine function, is given by d/dx (arcsin x) = 1/√(1 - x²). This function represents the angle whose sine is x and is only defined for values in the range -1 ≤ x ≤ 1. Differentiating arcsin(x) is essential in calculus, particularly in solving integrals and real-world applications. Various methods, such as the first principles approach, the chain rule, and the quotient rule, can be used to derive this formula systematically.

What is Derivative of arcsin x?

Derivative of arcsin function is denoted by d/dx(arcsin x) and its value is 1/√1-x². It returns the angle whose sine is a given number.

When the sine of y is equal to x:

sin y = x

Then the arcsine of x is equal to the inverse sine function of x, which is equal to y:

\(arcsin x = sin^{-1}x = y\)

Example: \(arcsin 1 = sin^{-1} 1 = π/2 rad = 90°\)

Graph of Arcsine: Arcsin x can be represented in graphical form as follows:

Values of Arcsin

Proof of Derivative of Arcsin x

We will learn how to differentiate arcsin x by using various differentiation rules:

1. Proof of Derivative of Arcsin by Quotient Rule
2. Proof of Derivative of Arcsin by first principle of derivative
3. Proof of Derivative of Arcsin by differentiating arcsin x using chain rule

Proof of Derivative of Arcsin by Quotient Rule

We can prove the derivative of arcsin by quotient rule using the following steps:

Step 1: Write sin y = x,

Step 2: Differentiate both sides of this equation with respect to x.

\(\begin{matrix}
{d\over{dx}}sin y = {d\over{dx}}x\\
cosy {d\over{dx}} y = 1
\end{matrix}\)

Step 3: Solve for \({dy\over{dx}}\)
\({d\over{dx}} y = {1\over{cosy}}\)

Step 4: Define cos y in terms of x using a reference triangle.

From the reference triangle, the adjacent side is \(\sqrt{(1 – x^2)}\) and the hypotenuse is 1. Thus, \(cos y = {\sqrt{(1 – x^2)}\over{1}}\) which means \({1\over{cosy}} = {1\over{\sqrt{(1 – x^2)}}}\)

Step 5: Substitute for cos y.

\({d\over{dx}} y = {1\over{cosy}} = {1\over{\sqrt{(1 – x^2)}}}\)

Step 6: Define arcsine.

Now we can define arcsine as:

\(y = sin^{-1}x\)

Step 7: Differentiate and write the result.

\({d\over{dx}}sin^{-1}x = {dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}\)

Proof of Derivative of Arcsin by Chain Rule

We can prove the derivative of arcsin by Chain rule using the following steps:

\(\begin{matrix}
\text{ Let } y = arcsin x\\
\text{ Taking sin on both sides, }\\
sin y = sin (arcsin x)\\
\text{ By the definition of an inverse function, we have, }\\
sin (arcsin x) = x\\
\text{ So the equation becomes }
sin y = x \\
\text{ Differentiating both sides with respect to x,}\\
{d\over{dx}} (sin y) = {d\over{dx}} (x)\\
cosy {d\over{dx}} y = 1\\
{d\over{dx}} y = {1\over{cosy}}\\
\text{ Using one of the trigonometric identities }\\
sin^y + cos^y = 1\\
{\therefore} cos y = \sqrt{1 – sin^2y} = \sqrt{1 – x^2}\\
{dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}\\
\text{ Substituting y = arcsin x back }\\
{d\over{dx}}(arcsin x) = arcsin’x = {1\over{\sqrt{1-x^2}}}
\end{matrix}\)

Proof of Derivative of Arcsin by First Principle

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

\(\begin{matrix}
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=arcsin x\\
f(x+h)=arcsin(x+h)\\
f(x+h)–f(x)= tan(x+h) – tan(x) = arcsin (x + h) – arcsin x\\
{f(x+h) – f(x)\over{h}}={ arcsin (x + h) – arcsin x\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {arcsin (x + h) – arcsin x\over{h}}\\
\text{ Assume that arcsin (x + h) = A and arcsin x = B }\\
sin A = x + h\\
sin B = x\\
sin A – sin B = (x + h) – x\\
sin A – sin B = h\\
If \text{ h → 0, (sin A – sin B) → 0 sin A → sin B or A → B or A – B → 0}\\
\lim _{A-B{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{A-B{\rightarrow}0} {(A – B)\over{(sin A – sin B)}}\\
\text{ sin A – sin B = 2 sin [(A – B)/2] cos [(A + B)/2] }\\
\lim _{A-B{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{A-B{\rightarrow}0} {(A – B)\over{[2 sin [(A – B)/2] cos [(A + B)/2]]}}\\
\text{ A – B → 0, we can have (A – B)/2 → 0 }\\
\lim _{{A-B\over{2}}{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{{A-B\over{2}}{\rightarrow}0}{1\over{(sin [(A – B)/2]\over{[(A – B)/2])}}} \lim _{A-B{\rightarrow}0} cos[(A + B)/2]\\
f’(x) = cos[(B + B)/2] = cos B\\
sin B = x\\
cos B = \sqrt{1 – sin^2B} = \sqrt{1 – x²}\\
\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {1\over{\sqrt{(1 – x^2)}}}\\
f’(x)={dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}
\end{matrix}\)

Properties of Arcsine

Solved Examples of Derivative of Arcsin x

Example 1:  What is the derivative of arcsin(x − 1)?

Solution: Derivative of inverse trigonometric functions. The general formula to differentiate the arcsin functions is

\(\begin{matrix}
\int{sin^{-1}u} = {1\over{\sqrt{(1 – u^2)}}} {du\over{dx}}\\
{d\over{dx}} sin^{-1}(x-1)={1\over{\sqrt{(1-(x-1)^2)}}}\times{d(x-1)\over{dx}}\\
{d\over{dx}} sin^{-1}(x-1)={1\over{\sqrt{(1-(x-1)^2)}}}
\end{matrix}\)

Example 2:  What is the derivative of arcsin(x\a)?

Solution:

To start off, let’s set this function equal to y
\(\begin{matrix}
y=sin^{−1}({x\over{a}})\\
siny=({x\over{a}})\\
\text{ Multiply a to both sides and taking the derivative }\\
{d\over{dx}}[asiny] = {d\over{dx}}\\
{dy\over{dx}} acosy = 1\\
{dy\over{dx}} = {1\over{acosy}}\\
\text{ Divide both sides to isolate }{dy\over{dx}}\\
{dy\over{dx}} = {1\over{a}}secy\\
secy = {1\over{cosy}}\\
\text{ from the image below}\\
secy = {a\over{\sqrt{a2^−x^2}}}\\
\text{ We will now substitute this value back into the answer for our derivative: }
{dy\over{dx}} = {1\over{a}}secy\\
{dy\over{dx}} = {1\over{a}}{a\over{\sqrt{a2^−x^2}}}\\
\text{ Canceling out the a in the numerator and denominator, we are left with our final answer: }\\
{dy\over{dx}} = {1\over{\sqrt{a2^−x^2}}}
\end{matrix}\)

Knowing the derivative of arcsin(x) is essential in solving complicated calculus problems, particularly in integration and real-world applications such as physics and engineering. Because the formula d/dx (arcsin x) = 1/√(1 - x²) is obtained through differentiation rules, understanding these methods improves problem-solving skills. Using the chain rule, quotient rule, or first principles, students can establish a solid basis in inverse trigonometric functions. Through practice, identifying when to apply this derivative will come naturally, making higher calculus topics more accessible and intuitive.

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