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Derivative of Arcsin x is d/dx(arcsin x) = 1/√1-x². It is denoted by d/dx(arcsin x) or d/dx(sin-1x). Derivative of Arcsin refers to the process of finding the rate of change in Arcsin x function with respect to the independent variable. Derivative of Arcsin x is also known as differentiation of Arcsin.
In this article, we will learn about the derivative of Arcsin and its formula including the proof of the formula using the first principle of derivatives, quotient rule, and chain rule method.
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Derivativeof a function is the rate of change of the function with respect to any independent variable. The derivative of a function f(x) is denoted as f'(x) or (d /dx)[f(x)]. The differentiation of a trigonometric function is called a derivative of the trigonometric function or trig derivatives. The derivative of a function f(x) is defined as:
f'(x0) = limh→0 [f(x0 + h) – f(x0)] / h
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Among the inverse trig derivatives, the derivative of the Arcsin x is one of the derivatives. Derivative of arcsin function represents the rate at which the arcsin curve is changing at a given point. It is denoted by d/dx(arcsin x) or d/dx(sin-1x). Arcsinx is also known as inverse sin x.
Derivative of the Arcsin x is 1/√1-x²
The formula for the derivative of Arcsin x is given by:
(d/dx) [Arcsin x] = 1/√1-x²
OR
(Arcsin x)’ = 1/√1-x²
Also Check, Inverse Trigonometric Function
The derivative of tan x can be proved using the following ways:
To prove derivative of Arcsin x by chain rule, we will use basic trigonometric and inverse trigonometric formula:
Here is the proof of derivative of Arcsin x:
Let y = arcsinx
Taking sin on both sides
siny = sin(arcsinx)
By the definition of an inverse function, we have,
sin(arcsinx) = x
So the equation becomes siny = x .....(1)
Differentiating both sides with respect to x,
d/dx (siny) = d/dx (x)
cosy · d/dx(y) = 1 [ As d/dx(sin x) = cos x]
dy/dx = 1/cosy
Using one of the trigonometric identities
sin2 y+cos2 y = 1
∴ cos y = √1 – sin2y = √1–x2 [From (1) we have siny = x]
dy/dx = 1/√(1–x2)
Substituting y = arcsin x
d/dx (arcsinx) = arcsin′x = 1/√1 - x2
Also Check, Chain Rule
To prove derivative of arcsin x using First Principle of Derivative, we will use basic limits and trigonometric formulas which are listed below:
We can prove the derivative of arcsin by First Principle using the following steps:
Let f(x) = arcsinx
By First principle we have
put f(x) = arcsinx, we get
....(1)
Assume that arcsin (x + h) = A and arcsin x = B
So we have,
sin A = x+h .....(2)
sin B = x .......(3)
Subtract (3) from (2), we have
sin A – sinB = (x+h) – x
sinA – sinB = h
If h → 0, (sin A – sin B) → 0
sin A → sin B or A → B
Substitute these values in eq(1)
Using sin A – sin B = 2 sin [(A – B)/2] cos [(A + B)/2], we get
which can be written as:
Now, we know limx→0 x/sinx = 1, therefore the above equation changes to
Using one of the trigonometric identities
sin2 y+cos2 y = 1
∴ cos B = √1 – sin2B = √1–x2 [Sin B = x from (3)]
f′(x) = dy/dx = 1 / √(1–x2)
Also, Check
Example 1: Find the derivative of y = arcsin (3x).
Solution:
Let f(x) = arcsin (3x).
We know that d/dx (arcsin x) = 1/√1 - x².
By chain rule,
d/dx(arcsin(3x)) = 1/√(1 - (3x)² · d/dx (3x)
= 1/ √(1 -9x²) · (3)
= 3/√(1 -9x²)
Hence, the derivative of y = arcsin (3x) is 3/√(1 -9x²).
Example 2: Find the derivative of y = arcsin (1/2x).
Solution:
Let f(x) = arcsin (1/2x).
We know that d/dx (arcsin x) = 1/√1 - x².
By chain rule,
d/dx(arcsin(1/2x)) = 1/√(1 - (1/2x)² · d/dx (1/2x)
= 1/ √(1 -(1/4x²) )· (-1/2x2)
= 1/√(4x2 - 1)/4x2 · (-1/2x2)
= -1/x√4x2 - 1
Hence, the derivative of y = arcsin (1/x) is -1/x√4x2 - 1.
Example 3: Find the derivative of y = x arcsin x.
Solution:
We have y = x arcsin x.
d/dx(arcsin(1/x)) = x · d/dx (arcsin x) + arcsin x · d/dx (x)
= x [1/√1-x²] + arcsin x (1)
= x/√1-x² + arcsin x
Hence, the derivative of y = arcsin (1/x) is x/√1-x² + arcsin x
1. Find the derivative of sin-1(5x).
2. Find the derivative of x3 sin-1(x).
3. Evaluate: d/dx [ sin-1(x) / x2 + 1 ]
4. Evaluate the derivative of sin-1(x) - tan(x)
To find the derivative of , we start by letting . This implies that . Differentiating both sides with respect to x using implicit differentiation, we get \sin(y) = x. Solving for , we have . Since and , it follows that . Thus, the derivative of is .
