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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoHass 15th EditionCh. 7 - Transcendental FunctionsProblema 7.6.31
Capitolo 7, Problema 7.6.31

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

Guida verificata passo dopo passo
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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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
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