   Chapter 3.2, Problem 15E ### 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

# Differentiate. y = t 3 + 3 t t 2 − 4 t + 3

To determine

To find: The differentiation of the function y=t3+3tt24t+3.

Explanation

Given:

The function y=t3+3tt24t+3.

Derivative rule:

(1) Quotient Rule: ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2(x)]2

(2) Constant multiple rule: ddx(cf)=cddx(f)

(3) Power Rule: ddx(xn)=nxn1

(4) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

(5) Difference rule: ddx(fg)=ddx(f)ddx(g)

Calculation:

The derivative of the function is dydx, which is obtained as follows.

dydt=ddt(y)=ddt(t3+3tt24t+3)

Use quotient rule (1) and simplify the terms,

dydx=(t24t+3)ddt[t3+3t](t3+3t)ddt[t24t+3][t24t+3]2

Apply the derivative rules (4), (5), (2), and (3) and simplify the terms,

dydt=(t24t+3)[ddt(t3)+ddt(3t)](t3+3t)[ddt(t2

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