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.

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

What is the Derivative of Sin Inverse x in Trigonometry?

The derivative of sin inverse x is 1/√(1-x²), where -1 < x < 1. Mathematically, it is written as d(sin^-1x )/ dx = 1/√(1-x²), -1 < x < 1

What is the Derivative of sin inverse x + cos inverse x?

The derivative of sin inverse x is 1/√(1-x²) and the derivative of cos inverse x is -1/√(1-x²). Therefore the derivative of sin inverse x + cos inverse x is 1/√(1-x²) + [-1/√(1-x²)] = 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 √(1-x²)?

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