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Ch. 3 - Derivatives
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 3, Problema 3.R.29

9–61. Evaluate and simplify y'.


y = tan^−1 √t²−1

Guida verificata passo dopo passo
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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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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