   Chapter 3.2, Problem 22E

Chapter
Section
Textbook Problem

Differentiate. V ( t ) = 4 + t t e t

To determine

To find: The differentiation of the function V(t)=4+ttet.

Explanation

Given:

The function V(t)=4+ttet.

Derivative rule:

(1) Quotient Rule: If f1(t) and f2(t) are both differentiable, then

ddt[f1(t)f2(t)]=f2(x)ddt[f1(t)]f1(t)ddt[f2(t)][f2(t)]2

(2) Product Rule: ddt[f1(t)f2(t)]=f1(t)ddt[f2(t)]+f2(t)ddt[f1(t)]

(3) Power Rule: ddt(tn)=ntn1

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

(5) Natural exponential function: ddt(et)=et

Calculation:

The derivative of the function V(t)=4+ttet is V(t), which is obtained as follows,

V(t)=ddt(V(t))=ddt(4+ttet)

Apply the quotient rule (1) and simplify the terms,

V(t)=[(tet)ddt(4+t)][(4+t)ddt(tet)](tet)2

Apply the sum rule (4) and simplify the terms,

V(t)=[(tet)(ddt(4)+ddt(t))][(4+t)ddt(tet)](tet)2 (1)

Obtain derivative ddt(tet).

Apply the product rule (2) and simplify the terms,

ddt(tet)=tddt(et)+etddt(t)

Apply the derivative rules (3) and (5),

ddt(tet)=t(et)+et(1)=tet+et

Thus, the derivative is ddt(tet)=tet+et

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