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-1x)' = 1/√(1-x2), that is, the derivative of sin inverse x is 1/√(1-x2). In other words, the rate of change of sin-1x at a particular angle is given by 1/√(1-x2), 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 anti-derivative of sin inverse x which is nothing but the integral of sin-1x.

1. What is the Derivative of Sin Inverse x? 2. Proof of Derivative of Sin Inverse x 3. Derivative of Sin Inverse x w.r.t. Cos Inverse √(1-x2) 4. Graph of Derivative of Sin Inverse of x 5. Anti-Derivative of Sin Inverse x 6. FAQs on Derivative of Sin Inverse x

The derivative of sin inverse x is 1/√(1-x2), 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/√(1-x2), therefore the graph of the derivative of sin inverse x will be the graph of 1/√(1-x2).

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/√(1-x2) 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-1x is:

d(sin-1x )/ dx = 1/√(1-x2), -1 < x < 1

We know that the derivative of sin inverse x is 1/√(1-x2), 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:

• cos2θ + sin2θ = 1
• Chain rule: (f(g(x)))' = f'(g(x)).g'(x)
• d(sin x)/dx = cos x

Assume y = sin-1x ⇒ 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 - sin2y) (Using cos2θ + sin2θ = 1)

⇒ dy/dx = 1/√(1 - x2) (Because sin y = x)

⇒ d(sin-1x)/dx = 1/√(1 - x2)

Hence, d(sin-1x)/dx = 1/√(1 - x2) , that is, derivative of sin inverse x is 1/√(1 - x2).

Now, that we know the derivative of sin inverse x is 1/√(1-x2), where -1 < x < 1, we will differentiate sin-1x with respect to another function, that is, cos-1√(1-x2). For this, we will assume cos-1√(1-x2) to be equal to some variable, say z, and then find the derivative of sin inverse x w.r.t. cos inverse √(1-x2).

Assume y = sin-1x ⇒ sin y = x

Using cos2θ + sin2θ = 1, we have cos θ = √(1 - sin2θ)

⇒ cos y = √(1 - sin2y) = √(1-x2)

Differentiating sin y = x w.r.t. x, we get

cos y (dy/dx) = 1

⇒ dy/dx = 1/cos y

⇒ dy/dx = 1/√(1-x2) - (1)

Assume z = cos-1√(1-x2) ⇒ sin z = x and cos z = √(1-x2) (Using cos2θ + sin2θ = 1)

Now, differentiating cos z = √(1-x2) w.r.t. x, we have

-sin z (dz/dx) = -2x/2√(1-x2)

⇒ -x(dz/dx) = -x/√(1-x2)

⇒ dz/dx = 1/√(1-x2)

⇒ dx/dz = √(1-x2) - (2)

Now, we need to determine the value of d(sin-1x)/d(cos-1√(1-x2)) = dy/dz

dy/dz = dy/dx × dx/dz

= [1/√(1-x2)] × √(1-x2)

= 1

Hence, d(sin-1x)/d(cos-1√(1-x2)) = 1, that is, derivative of sin inverse x w.r.t. cos inverse √(1-x2) is 1.

As the derivative of sin inverse x is 1/√(1-x2), -1 < x < 1, therefore the graph of the derivative of sin inverse x is similar to the graph of the function 1/√(1-x2) such that x lies between -1 and 1. First, let us see how the graphs of sin-1x and the derivative of sin inverse x look like.

The anti-derivative of sin inverse x is nothing but the integral of sin inverse x. As the name suggests, anti-derivative is the inverse process of differentiation. The anti-derivative of sin inverse x is x sin-1x + √(1-x2) + C, where C is the constant of integration. Hence, we have obtained the anti-derivative of sin inverse x and sin x + C.

∫sin-1 x = x sin-1x + √(1-x2) + C

Related Topics on Derivative of Sin Inverse x

• 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/√(1-x2), where -1 < x < 1
• d(sin-1x)/d(cos-1√(1-x2)) = 1, that is, derivative of sin inverse x w.r.t. cos inverse √(1-x2) is 1.
• ∫sin-1 x = x sin-1x + √(1-x2) + C