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Problem 61 Hard Difficulty

(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}. $

Answer

(a) $=f^{\prime} g h+f g^{\prime} h+f g h^{\prime}$
(b) $=3[f(x)]^{2} f^{\prime}(x)$
(c) $=3 e^{3 x}$

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Video Transcript

he has clear. So when you right here. So for part Amy of you is equal to F G, then that makes the derivative of you beagle to the derivative of half times G plus tough times the derivative of gene. We've the product roll, so f g h this equal to you have a JJ. So when we derive this, this is equal to the derivative of U H, which is equal to the derivative of you terms H plus U turns the derivative of beach. So we got the derivative of F g H be equal to the derivative of afternoon's G plus F tongues, the derivative of G terms each plus G h Did they rip it if which is equal to the derivative of F tongues G H plus f times the derivative of G times H plus F G in the derivative of h, your part B, we're gonna make f B equal to G and be equal to age. And we have to formula in part a. And this gives us of cubed the derivative this equal to the first derivative of Times Square. Let's, uh, times the derivative of that times F plus F Square turns the derivative of which is equal to three terms, a derivative of of terms of square. In other words, do you over defects ah x You is equal to three times off Max square tongues the derivative after after backs. For part C, we're gonna use part B. So we have Why is equal to he to the three X, which is equal to e to the X cubed. This is the dirt formula for the derivative of a function cube. So we make up of X B equal to eat the X, and that makes the derivative. We also eat the axe the A D over D backs per eat the ex cute which is equal to D over t x f x cubed, which is people 23 turns e the X square turned Eat the ax when we go three me to the three x