Derivative of Sin Inverse x
The differentiation of sin inverse x is the process of evaluating the derivative of sin inverse x or determining the rate of change of sin inverse x with respect to the variable x. The derivative of the sine inverse function is written as (sin^{1}x)' = 1/√(1x^{2}), that is, the derivative of sin inverse x is 1/√(1x^{2}). In other words, the rate of change of sin^{1}x at a particular angle is given by 1/√(1x^{2}), where 1 < x < 1. Inverse trigonometric functions are used in different fields of engineering, physics, navigation, and geometry.
Now, the derivative of sin inverse x can be calculated using different methods. It can be derived using the limits definition and inverse function theorem. In this article, we will calculate the derivative of sin inverse x and also discuss the antiderivative of sin inverse x which is nothing but the integral of sin^{1}x.
What is the Derivative of Sin Inverse x?
The derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1. Derivatives of all inverse trigonometric functions can be calculated using the method of implicit differentiation. The derivative of a function characterizes the rate of change of the function at some point. The process of finding the derivative is called differentiation. The differentiation of sin inverse x can be done in different ways and it can be derived using the definition of the limit, and inverse function theorem. Since the derivative of sin inverse x is 1/√(1x^{2}), therefore the graph of the derivative of sin inverse x will be the graph of 1/√(1x^{2}).
Derivative of Sin Inverse x Formula
Now, we will write the derivative of sin inverse x mathematically. The derivative of a function is the slope of the tangent to the function at the point of contact. Hence, 1/√(1x^{2}) is the slope function of the tangent to the graph of sin inverse x at the point of contact. Mostly, we memorize the derivative of sin inverse x. An easy way to do that is knowing the fact that the derivative of sin inverse x is the negative of the derivative of cos inverse x and the derivative of cos inverse x is the negative of the derivative of sin inverse x. The mathematical expression to write the differentiation of sin^{1}x is:
d(sin^{1}x )/ dx = 1/√(1x^{2}), 1 < x < 1
Proof of Derivative of Sin Inverse x
We know that the derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1. Now, we will derive the derivative using some differentiation formulas. To derive the derivative of sin inverse x, we will use the following formulas:
 cos^{2}θ + sin^{2}θ = 1
 Chain rule: (f(g(x)))' = f'(g(x)).g'(x)
 d(sin x)/dx = cos x
Assume y = sin^{1}x ⇒ sin y = x
Differentiating both sides of sin y = x w.r.t. x, we have
cos y dy/dx = 1
⇒ dydx = 1/cos y
⇒ dy/dx = 1/√(1  sin^{2}y) (Using cos^{2}θ + sin^{2}θ = 1)
⇒ dy/dx = 1/√(1  x^{2}) (Because sin y = x)
⇒ d(sin^{1}x)/dx = 1/√(1  x^{2})
Hence, d(sin^{1}x)/dx = 1/√(1  x^{2}) , that is, derivative of sin inverse x is 1/√(1  x^{2}).
Derivative of Sin Inverse x w.r.t. Cos Inverse √(1x^{2})
Now, that we know the derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1, we will differentiate sin^{1}x with respect to another function, that is, cos^{1}√(1x^{2}). For this, we will assume cos^{1}√(1x^{2}) to be equal to some variable, say z, and then find the derivative of sin inverse x w.r.t. cos inverse √(1x^{2}).
Assume y = sin^{1}x ⇒ sin y = x
Using cos^{2}θ + sin^{2}θ = 1, we have cos θ = √(1  sin^{2}θ)
⇒ cos y = √(1  sin^{2}y) = √(1x^{2})
Differentiating sin y = x w.r.t. x, we get
cos y (dy/dx) = 1
⇒ dy/dx = 1/cos y
⇒ dy/dx = 1/√(1x^{2})  (1)
Assume z = cos^{1}√(1x^{2}) ⇒ sin z = x and cos z = √(1x^{2}) (Using cos^{2}θ + sin^{2}θ = 1)
Now, differentiating cos z = √(1x^{2}) w.r.t. x, we have
sin z (dz/dx) = 2x/2√(1x^{2})
⇒ x(dz/dx) = x/√(1x^{2})
⇒ dz/dx = 1/√(1x^{2})
⇒ dx/dz = √(1x^{2})  (2)
Now, we need to determine the value of d(sin^{1}x)/d(cos^{1}√(1x^{2})) = dy/dz
dy/dz = dy/dx × dx/dz
= [1/√(1x^{2})] × √(1x^{2})
= 1
Hence, d(sin^{1}x)/d(cos^{1}√(1x^{2})) = 1, that is, derivative of sin inverse x w.r.t. cos inverse √(1x^{2}) is 1.
Graph of Derivative of Sin Inverse of x
As the derivative of sin inverse x is 1/√(1x^{2}), 1 < x < 1, therefore the graph of the derivative of sin inverse x is similar to the graph of the function 1/√(1x^{2}) such that x lies between 1 and 1. First, let us see how the graphs of sin^{1}x and the derivative of sin inverse x look like.
AntiDerivative of Sin Inverse x
The antiderivative of sin inverse x is nothing but the integral of sin inverse x. As the name suggests, antiderivative is the inverse process of differentiation. The antiderivative of sin inverse x is x sin^{1}x + √(1x^{2}) + C, where C is the constant of integration. Hence, we have obtained the antiderivative of sin inverse x and sin x + C.
∫sin^{1 }x = x sin^{1}x + √(1x^{2}) + C
Related Topics on Derivative of Sin Inverse x
 Inverse Trigonometric Formulas
 Inverse Trig Functions Calculator
 Cos Inverse Formula
 Inverse Trigonometric Functions
Important Notes on Derivative of Sin Inverse x
 The derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1
 d(sin^{1}x)/d(cos^{1}√(1x^{2})) = 1, that is, derivative of sin inverse x w.r.t. cos inverse √(1x^{2}) is 1.
 ∫sin^{1 }x = x sin^{1}x + √(1x^{2}) + C
Examples on Derivative of Sin Inverse x

Example 1: Use the derivative of sin inverse x formula to determine the derivative of sin^{1}(x^{3}).
Solution: The derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1. To determine the derivative of sin^{1}(x^{3}), we will use the chain rule method.
d(sin^{1}(x^{3}))/dx = 1/√[1(x^{3})^{2}] × 3x^{2}
= 3x^{2}/(1x^{6})
Answer: d(sin^{1}(x^{3}))/dx = 3x^{2}/(1x^{6})

Example 2: Is the derivative of sin inverse x equal to the derivative of negative sin inverse x?
Solution: The derivative of sin inverse x is 1/√(1x^{2}). The derivative of negative sin inverse x is equal to the negative of the derivative of sin inverse x, that is, negative of 1/√(1x^{2}).
Hence the derivative of sin^{1}x is (1/√(1x^{2})) = 1/√(1x^{2})
Answer: No, d(sin^{1}x)/dx = 1/√(1x^{2}), 1 < x < 1
FAQs on Derivative of Sin Inverse x
What is the Derivative of Sin Inverse x in Trigonometry?
The derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1. Mathematically, it is written as d(sin^{1}x )/ dx = 1/√(1x^{2}), 1 < x < 1
What is the Derivative of sin inverse x + cos inverse x?
The derivative of sin inverse x is 1/√(1x^{2}) and the derivative of cos inverse x is 1/√(1x^{2}). Therefore the derivative of sin inverse x + cos inverse x is 1/√(1x^{2}) + [1/√(1x^{2})] = 0. Hence, the derivative of sin inverse x + cos inverse x is zero.
What is the Derivative of sin inverse x with respect to cos inverse √(1x^{2})?
The derivative of sin inverse x w.r.t. cos inverse √(1x^{2}) is 1, that is, d(sin^{1}x)/d(cos^{1}√(1x^{2})) = 1.
How to Find the Derivative of sin inverse x?
The derivative of sin inverse x can be derived using the definition of the limits, inverse function theorem and the method of implicit differentiation. The derivative of sin inverse x is 1/√(1x^{2}), where 1 < x < 1