   Chapter 3.11, Problem 39E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the derivative. Simplify where possible.39. g ( t ) = t coth t 2 + 1

To determine

To find: The derivative of the function.

Explanation

Given:

The function g(t)=tcott2+1.

Derivative rules used:

(1) The chain rule: dydx=dydududx

(2) The derivative of the hyperbolic function is, ddu(cothu)=csch2u.

Calculation:

The derivative of the function g(t)=tcott2+1 is computed as follows,

g(t)=ddt(tcott2+1)

Apply the product rule and simplify the terms,

g(t)=tddt(cott2+1)+(cott2+1)ddt(t)=tddt(cott2+1)+(cott2+1)(1)

Let u=t2+1.

g(t)=tddt(cothu)+cothu

Apply the chain rule and simplify the terms,

g(t)=tddu(cothu)dudt+cothu=t(csch2u)dudt+cothu      (Qddu(cothu)=csch2u)=</

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