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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

If f ( t ) = 4 t + 1 find f″(2).

To determine

To find: The value of f(2).

Explanation

Given:

The function f(t)=4t+1.

Derivative rule:

(1) The Power Rule combined with the Chain Rule:

If n is any real number and g(x) is differentiable function, then

ddx[g(x)]n=n[g(x)]n1g(x) (1)

(2) Quotient Rule:

 If f1(x) and f2(x) are both differentiable, then ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

Calculation:

The first derivative of f(t) is f(t), which is obtained as follows,

f(t)=ddt(f(x)) =ddt(4t+1)=ddt(4t+1)12

Apply the power rule combined with the chain rule(1),

f(t)=12(4t+1)121ddt(4t+1)=12(4t+1)122(4(1)+0)=12(4t+1)12(4)=24t+1

Thus, the first derivative of the function f(t)=4t+1 is f(t)=24t+1.

The second derivative of f(t) is f(t), which is obtained as follows

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