Question

Find the derivative. $$y=\sqrt{1-t^{2}}+\sin ^{-1} t$$

   Find the derivative.
$$y=\sqrt{1-t^{2}}+\sin ^{-1} t$$

Calculus for AP

Calculus for AP

Jon Rogawski & Colin… 3rd Edition

Chapter 3, Problem 39

Instant Answer

Solved by Expert Darshan Maheshwari

10/20/2020

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We have to find the derivative of this function. The function is a sum of two functions, so we can find the derivative of each function separately and then add them up. Show more…

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Find the derivative. $$y=\sqrt{1-t^{2}}+\sin ^{-1} t$$

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Key Concepts

Differentiation of Inverse Trigonometric Functions

The derivative of inverse trigonometric functions is an important aspect of calculus. Specifically, for the inverse sine function, sin?¹(t), the derivative is 1/sqrt(1-t²). This result comes from the geometric interpretation of arcsine and is used when differentiating functions that include inverse trigonometric expressions.

Chain Rule

The chain rule is a fundamental technique in differentiation that allows one to compute the derivative of composite functions. When a function is written as f(g(t)), its derivative is given by f'(g(t)) multiplied by g'(t). This rule is essential when the function inside another function (an inner function) needs to be differentiated alongside the outer function.

Differentiation of Square Root Functions

Differentiating a square root function involves rewriting the square root as an exponent of 1/2 and then applying the power rule in combination with the chain rule. For example, if a function is in the form sqrt(u(t)), it can be expressed as [u(t)]^(1/2), and its derivative is (1/(2*sqrt(u(t)))) times the derivative of u(t).

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