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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 3, Problem 3.R.29

9–61. Evaluate and simplify y'.


y = tan^−1 √t²−1

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1
First, identify the function y in terms of t. Here, y = tan^(-1)(√(t²−1)).
To find y', the derivative of y with respect to t, use the chain rule. The chain rule states that if a function y = f(g(t)), then y' = f'(g(t)) * g'(t).
Recognize that y = tan^(-1)(u) where u = √(t²−1). The derivative of tan^(-1)(u) with respect to u is 1/(1+u²).
Next, find the derivative of u = √(t²−1) with respect to t. Use the chain rule again: if u = (t²−1)^(1/2), then u' = (1/2)(t²−1)^(-1/2) * 2t = t/√(t²−1).
Combine the derivatives using the chain rule: y' = (1/(1+u²)) * (t/√(t²−1)). Substitute u = √(t²−1) back into the expression to simplify y'.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of Inverse Functions

The derivative of an inverse function, such as the arctangent function, can be found using the formula: if y = tan^(-1)(u), then dy/dx = 1/(1 + u^2) * du/dx. This concept is essential for differentiating functions that involve inverse trigonometric functions.
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Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. If a function y = f(g(t)), the derivative is given by dy/dt = f'(g(t)) * g'(t). This rule is crucial for evaluating the derivative of y = tan^(-1)(√(t² - 1)) since it involves a composition of functions.
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Simplifying Derivatives

After finding the derivative, simplifying the expression is often necessary to make it more manageable or interpretable. This may involve factoring, combining like terms, or rationalizing the denominator. Understanding how to simplify derivatives is key to presenting the final answer clearly.
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