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Numerade Educator



Problem 32 Easy Difficulty

Suppose $ f(\pi/3) = 4 $ and $ f'(\pi/3) = -2, $ and let $ g(x) = f(x) \sin x $ and $ h(x) = (\cos x)/f(x). $ Find
(a) $ g'(\pi/3) $
(b) $ h'(\pi/3) $


a) $g^{\prime}\left(\frac{\pi}{3}\right)=2-\sqrt{3}$
b) $h^{\prime}(\pi / 3)=\frac{-2 \sqrt{3}+1}{16}$

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Video Transcript

Hey, it's clear. So when you read here so we have G of X is equal to f of X sign X. We take the derivative his equals of X using the product rule. It's the derivative of sign which just go sign close signed times the derivative of F F ax. We're going to substitute X is equal to pi thirds we get to minus 33 square root of three. For part B, you have each of eggs is equal to co sign over f of X. We take the derivatives using the quotient rule, which is equal to off our backs times the derivative of co sign of X, which is negative. Sign of X minus co sign of X times the derivative of flux all over f of X square. When we're gonna plug in pine thirds, we get negative too squared of three plus one over 16