Derivative of arcsin(x) derivation
Derivative of arcsin(x) derivation

Overview

Test Series

Derivative of arcsin x is 1/√1-x². It is written as d/dx(arcsin x) = 1/√1-x². Arcsin function is the inverse of the sine function and is a pure trigonometric function. We will learn how to differentiate arcsin x by using various differentiation rules like the first principle of derivative, differentiate arcsin x using chain rule and differentiate arcsin x using the quotient rule. Arcsin of x is defined as the inverse sine function of x when -1 ≤ x ≤ 1.

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

 x arcsin(x) (rad) arcsin(x) (°) -1 -π/2 -90° $$-\sqrt3\over2$$ -π/3 -60° $$-\sqrt2\over2$$ -π/4 -45° $$-1\over2$$ -π/6 -30° 0 0 0° $$1\over2$$ π/6 30° $$\sqrt2\over2$$ π/4 45° $$\sqrt3\over2$$ π/3 60° 1 π/2 90°

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

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.

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

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

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

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

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

 Rule name Rule Sine of arcsine sin( arcsin x ) = x Arcsine of sine arcsin( sin x ) = x+2kπ, when k∈ℤ (k is integer) Arcsin of negative argument arcsin(-x) = – arcsin x Complementary angles arcsin x = π/2 – arccos x = 90° – arccos x Arcsin sum $$arcsin\alpha + arcsin\beta = arcsin(\alpha\sqrt{(1-\beta^2)} + \beta\sqrt{(1-\alpha^2)})$$ Arcsin difference $$arcsin\alpha – arcsin\beta = arcsin(\alpha\sqrt{(1-\beta^2)} – \beta\sqrt{(1-\alpha^2)})$$ Cosine of arcsine $$cos(arcsin x) = sin(arccosx) = \sqrt{1-x^2}$$ Tangent of arcsine $$tan(arcsin x) = {x\over{\sqrt{1-x^2}}}$$ Derivative of arcsine $${d\over{dx}}(arcsin x) = arcsin’x = {1\over{\sqrt{1-x^2}}}$$ Indefinite integral of arcsine $$\int(arcsin x) dx = x(arcsin x) + {\sqrt{1-x^2}} + C$$ Corollary $$\dfrac {d {\arcsin {\frac x a} } } {d x} = \dfrac 1 {\sqrt {a^2 – x^2} }$$

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

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