Derivative of Sin inverse x

In mathematics, Differentiation is one of the two most important and useful concepts in calculus, along with integration. Differentiation is a method for evaluating the derivative of a function. Differentiation is a mathematical process for calculating a function’s instantaneous rate of change based on one of its variables. The most common example is velocity, which is defined as the rate at which a displacement changes over time. Finding a derivative is the inverse of anti-differentiation.

If x is one variable and y is another variable, then dy/dx is the rate of change of x with respect to y. This is the general expression for a function’s derivative, and it is written as f'(x) = dy/dx, with y = f(x) being any function.

The process of calculating the derivative of sin inverse x or finding the rate of change of sin inverse x with regard to the variable x is known as the differentiation of sin inverse x. The sine inverse function’s derivative is expressed as (sin-1 x)’ = 1/ √ (1-x2), i.e., the derivative of sin inverse x is 1/ √ (1-x2) . In other words, at a given angle, the rate of change of sin-1x is given by 1/ √ (1-x2), where -1< x < 1. Inverse trigonometric functions are utilized in engineering, physics, navigation, and geometry, among other subjects.

Different approaches can now be used to determine the derivative of sin inverse x. The inverse function theorem and the definition of the limit can be used to derive it.

The method of implicit differentiation can be used to calculate derivatives of all inverse trigonometric functions. The rate of change of a function at a given point is defined by its derivative. Differentiation is the process of finding the derivative. Differentiation of sin inverse x can be done in a variety of methods, including using the limit definition and the inverse function theorem. The graph of the derivative of sin inverse x will equal the graph of 1/√(1-x2) since the derivative of sin inverse x is 1/√(1-x2).

• Differentiation of Sin Inverse x 

The derivative of sin inverse x will now be written formally. The slope of the tangent to the function at the point of contact is the derivative of a function. As a result, at the point of contact, 1/√(1-x2) is the slope function of the tangent to the graph of sin inverse x. We memorize the derivative of sin inverse x most of the time. Knowing that the derivative of sin inverse x is the negative of the derivative of cos inverse x and that the derivative of cos inverse x is the negative of the derivative of sin inverse x provides a simple way to do this. The differentiation of sin-1 x is expressed mathematically as:

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

• Proof:

The derivative of sin inverse x is 1/√(1-x2), where -1< x < 1 is known. We’ll now use some differentiation formulas to calculate the derivative. We’ll use the following formulas to find the derivative of sin inverse x:

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

Let y = sin-1x 

Then,  sin y = x

Both sides of sin y = x is differentiated w.r.t. x, we get

cosy dy/dx = 1

⇒ dydx = 1/cos y

⇒ dy/dx = 1/√(1 – sin2y) (Use cos2θ + sin2θ = 1)

⇒ dy/dx = 1/√(1 – x2) (since sin y = x)

⇒ d(sin-1x)/dx = 1/√(1 – x2)

Thus, 

d(sin-1x)/dx = 1/√(1 – x2), which means derivative of sin inverse x is 1/√(1 – x2).

Derivative of Sin Inverse x w.r.t. Cos Inverse √(1-x2):

• We’ll differentiate sin-1 x with regard to another function, cos-1√(1-x2), now that we know the derivative of sin inverse x is 1/√(1-x2), where -1 < x < 1. To do so, we’ll set cos-1√(1-x2) equal to some variable, say z, and then determine the derivative of sin inverse x with respect to cos-1√(1-x2).
• Let y = sin-1x which gives sin y = x
• By using the identity cos2θ + sin2θ = 1, we get cos θ = √(1 – sin2θ)
• ⇒ cos y = √(1 – sin2y) = √(1-x2)
• On Differentiating sin y = x w.r.t. x, we have
• cos y (dy/dx) = 1
• ⇒ dy/dx = 1/cos y
• ⇒ dy/dx = 1/√(1-x2) —- (1)
• Let z = cos-1√(1-x2) 
• ⇒ sin z = x and cos z = √(1-x2) (Use cos2θ + sin2θ = 1)
• On differentiating cos z = √(1-x2) w.r.t. x, we get
• -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
• Thus, d(sin-1x)/d(cos-1√(1-x2)) = 1, which means, derivative of sin inverse x w.r.t. cos inverse √(1-x2) is 1. 

Conclusion:

The derivative of sin inverse x is expressed as d(sin-1 x)/dx and represents the rate of change of sin inverse x with regard to x. The differentiation of sin inverse x is also the process of getting the derivative of sin inverse x. Substitution, the first principle, and other approaches can be used to find the derivative of sin-1 x.