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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 7 - Transcendental FunctionsProblem 7.6.31
Chapter 7, Problem 7.6.31

In Exercises 21–48, find the derivative of y with respect to the appropriate variable.
31. y=arccot(√t)

Verified step by step guidance
1
Identify the function to differentiate: \(y = \arccot(\sqrt{t})\). We need to find \(\frac{dy}{dt}\).
Recall the derivative formula for \(y = \arccot(u)\) with respect to \(u\): \(\frac{dy}{du} = -\frac{1}{1 + u^2}\).
Set \(u = \sqrt{t} = t^{1/2}\). Then find \(\frac{du}{dt}\) using the power rule: \(\frac{du}{dt} = \frac{1}{2} t^{-1/2}\).
Apply the chain rule: \(\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = -\frac{1}{1 + u^2} \cdot \frac{1}{2} t^{-1/2}\).
Substitute back \(u = \sqrt{t}\) into the expression to write the derivative entirely in terms of \(t\).

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Key Concepts

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

Derivative of Inverse Trigonometric Functions

Inverse trigonometric functions like arccot(x) have specific derivative formulas. For arccot(x), the derivative with respect to x is -1/(1 + x²). Understanding these derivatives is essential to differentiate expressions involving inverse trig functions.
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Derivatives of Other Inverse Trigonometric Functions

Chain Rule

The chain rule is used to differentiate composite functions. When y = arccot(√t), you first differentiate arccot(u) with respect to u, then multiply by the derivative of u = √t with respect to t. This rule allows handling nested functions effectively.
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Intro to the Chain Rule

Derivative of Square Root Functions

The square root function √t can be rewritten as t^(1/2). Its derivative with respect to t is (1/2)t^(-1/2). Knowing how to differentiate root functions is necessary when applying the chain rule to expressions like arccot(√t).
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Root Test
Related Practice
Textbook Question

Each of Exercises 25–36 gives a formula for a function y=f(x). In each case, find f^(-1)(x) and identify the domain and range of f^(-1). As a check, show that f(f^(-1)(x))=f^(-1)(f(x))=x.


f(x) = (x + b) / (x − 2), b > −2 and constant

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