   Chapter 2.3, Problem 87E

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# (a) Use the Product Rule twice to prove that if f, g, and h are differentiable, then ( f g h ) ′ = f ′ g h + f g ′ h + f g h ′ .(b) Taking f = g = h in part (a), show that d d x [ f ( x ) ] 3 = 3 [ f ( x ) ] 2 f ′ ( x ) (c) Use part (b) to differentiate y = ( x 4 + 3 x 3 + 17 x + 82 ) 3 .

To determine

(a)

To prove:  f g h '=f'g h+f g'h+f g h'

Explanation

1) Concept:

Product rule: If u and v are both differentiable, then

ddxuxvx=uxddxvx+vxddxux

2) Calculations:

Assume u = fg and v = h

Then using the product rule,

ddx(fg)(h)=(fg)ddxh+hddxfg

ddx(fg)(h)=fgh'+h*(fg)'

Now to find (fg)' again use the product rule.

So fg'= ddxfg becomes,

ddxf*g=(f)ddxg+gddx

To determine

(b)

To show:

ddxfx3=3 fx2*f'x

To determine

(c)

To differentiate: y=x4+3x3+17x+823

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