SAT Derivative of Arcsin x Formula & Proof by Quotient Rule, Chain Rule & First Principle

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:

Example: 

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

Values of Arcsin

x	arcsin(x) (rad)	arcsin(x) (°)
-1	-π/2	-90°
	-π/3	-60°
	-π/4	-45°
	-π/6	-30°
0	0	0°
	π/6	30°
	π/4	45°
	π/3	60°
1	π/2	90°

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

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

From the reference triangle, the adjacent side is and the hypotenuse is 1. Thus, which means

Step 5: Substitute for cos y.

Step 6: Define arcsine.

Now we can define arcsine as:

Step 7: Differentiate and write the result.

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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