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Numerade Educator



Problem 43 Hard Difficulty

Suppose that $ f(5) = 1, f' (5) = 6, g(5) = -3, $ and $ g'(5) = 2. $ Find the following values.
(a) $ (fg)' (5) $

(b) $ (f/g)' (5) $

(c) $ (g/f)' (5) $


(a) -16
(b) -20 / 9
(c) 20

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Video Transcript

feats. Clearasil on rain here. So, for part A, we're gonna use the product rule. Let me get F G. Derivative five is equal to derivative of F five GF five less five derivative of G of five. You plug it in and you get six times negative. Three plus one times two, which is equal to negative 16. We have part being f departed by G. The derivative of five is equal to the drill bit it, uh, the five g of five minus F five. The derivative of chief of Fly all over G of five square and you get six times negative. Three minus one times two over a negative three square, giving us negative. 20 over nine For part C, you have Jeanne divided by s the derivative five, which is equal to the derivative of G of five. Yes, a five minus g of five. The derivative of F of five all over five square. This is equal to two time one minus negative. Three. I'm six over one square, which is equal to 20