Derivatives of Inverse Trigonometric Functions: Understanding sin-1(1-t)

This article explains the process of finding the derivative of the function ( y sin^{-1}(1-t) ) using the chain rule and the derivative of the inverse sine function. Understanding this concept is crucial for advanced calculus and can be applied to various practical problems in science and engineering.

1. Introduction to the Function

The function we are dealing with is ( y sin^{-1}(1-t) ). The goal is to find the derivative ( frac{dy}{dt} ).

2. Deriving the Derivative Using the Chain Rule

The derivative of the inverse sine function ( sin^{-1}x ) is given by:

[ frac{d}{dx} sin^{-1}x frac{1}{sqrt{1 - x^2}} ]

In the given function, ( x 1 - t ). Therefore, we need to apply the chain rule:

[ frac{dy}{dt} frac{d}{dx} sin^{-1}(1-t) cdot frac{dx}{dt} ]

3. Calculating ( frac{dx}{dt} )

First, let's compute ( frac{dx}{dt} ):

[ frac{d}{dt}(1 - t) -1 ]

4. Applying the Chain Rule

Substituting ( frac{dx}{dt} -1 ) and the derivative of ( sin^{-1}x ) into the chain rule, we get:

[ frac{dy}{dt} frac{1}{sqrt{1 - (1 - t)^2}} cdot (-1) ]

5. Simplifying the Expression

Let's simplify the expression inside the square root:

[ 1 - (1 - t)^2 1 - (1 - 2t t^2) 1 - 1 2t - t^2 2t - t^2 ]

Thus, we have:

[ frac{dy}{dt} -frac{1}{sqrt{2t - t^2}} ]

This result is valid for values of ( t ) such that ( 2t - t^2 geq 0 ), which corresponds to ( 0 leq t leq 2 ).

6. Alternative Method Using Trigonometric Identities

Another way to derive the same result is by using trigonometric identities. Starting from ( y sin^{-1}(1-t) ), we have:

[ sin y 1 - t ]

By taking the derivative of both sides and using the chain rule on the left-hand side (LHS), we obtain:

[ cos y cdot frac{dy}{dt} -1 ]

Using the identity ( cos^2 y sin^2 y 1 ), we can rearrange it to get:

[ cos y sqrt{1 - sin^2 y} ]

Substituting this into the equation, we get:

[ sqrt{1 - sin^2 y} cdot frac{dy}{dt} -1 ]

Rearranging this, we find:

[ frac{dy}{dt} -frac{1}{sqrt{1 - sin^2 y}} ]

Substituting back ( sin y 1 - t ), we obtain:

[ frac{d}{dt} sin^{-1}(1-t) -frac{1}{sqrt{1 - (1-t)^2}} ]

Following the same simplification as before, we get the same result:

[ frac{dy}{dt} -frac{1}{sqrt{2t - t^2}} ]

7. Conclusion

The derivative of ( y sin^{-1}(1-t) ) is ( frac{dy}{dt} -frac{1}{sqrt{2t - t^2}} ) for ( 0 leq t leq 2 ).