Question

(a) Use the Product Rule twice to prove that it $ f,g, $ and $ h $ are differentiable. then $ (fgh)' = f'gh + fgh'. + fgh'. $ (b) Taking $ f = g = h $ in part (a), show that $ frac {d}{dx}[f(x)]^3=3[f(x)]^2f'(x) $ (c) Use part (b) to differentiate $ y = e^{3x}. $

Name: a use the product rule twice to prove that it fg and h are differentiable then fgh fgh fgh fgh b taking f g h in part a show that frac ddxfx33fx2fx c use part b to differentiate y e3x 2
Uploaded: 2019-12-06T06:53:50-08:00
Duration: 4 min 10 s
Channel: Clarissa Noh
Description: a use the product rule twice to prove that it fg and h are differentiable then fgh fgh fgh fgh b taking f g h in part a show that frac ddxfx33fx2fx c use part b to differentiate y e3x 2

          (a) Use the Product Rule twice to prove that it $ f,g, $ and $ h $ are differentiable. then $ (fgh)' = f'gh + fgh'. + fgh'. $ 
(b) Taking $ f = g = h $ in part (a), show that

$ frac {d}{dx}[f(x)]^3=3[f(x)]^2f'(x) $

(c) Use part (b) to differentiate $ y = e^{3x}. $

Added by Daisy J.

Calculus Early Transcendentals

James Stewart 6th Edition

Chapter 3

Instant Answer

Solved by Expert Clarissa Noh

12/06/2019

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Step 1: Use the Product Rule twice to prove that if $f$, $g$, and $h$ are differentiable, then $(fgh)' = f'gh + fg'h + fgh'$. Show more…

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(a) Use the Product Rule twice to prove that it $ f,g, $ and $ h $ are differentiable. then $ (fgh)' = f'gh + fgh'. + fgh'. $ (b) Taking $ f = g = h $ in part (a), show that $ frac {d}{dx}[f(x)]^3=3[f(x)]^2f'(x) $ (c) Use part (b) to differentiate $ y = e^{3x}. $