   Chapter 3.2, Problem 61E

Chapter
Section
Textbook Problem

(a) Use the Product Rule twice to prove that if f, g, and h are differentiable, then (fgh)' = f'gh + fg'h + fgh'.(b) Taking f = g = h in part (a), show that d x [ f ( x ) ] 3 = 3 [ f ( x ) ] 2 f ' ( x ) (c) Use part (b) to differentiate y = e3x.

(a)

To determine

To show: If f,g and h are differentiable then (fgh)=fgh+fgh+fgh.

Explanation

Derivative rule:

Product Rule: ddx[f1(x)f2(x)]=f1(x)ddx[f2(x)]+f2(x)ddx[f1(x)]

Proof:

Let p=fg.

The derivative of p=fg is computed as follows,

p=(fg)

Apply the product rule and simplify,

p=fd(g)+gd(f)=fg+gf

Thus, the derivative of p=fg is p=fg+gf

(b)

To determine

To show: ddx[f(x)]3=3[f(x)]2f(x)

(c)

To determine

To find: Differentiation of the function.

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