Overview
Test Series
The 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.
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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:
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° |
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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