Problem

If $ f $ is a differentiable function, find an ex…

06:56

Problem 51 Hard Difficulty

If $ g $ is a differentiable function, find ail expression tor the derivative of each of the following functions.

(a) $ y = xg(x) $

(b) $ y = \frac {x}{g(x)} $

(c) $ y = \frac {g(x)}{x} $

Answer

(a) $y^{\prime}=x g^{\prime}(x)+g(x)$
(b) $y^{\prime}=\frac{g(x)-x \cdot g^{\prime}(x)}{[g(x)]^{2}}$
(c) $y^{\prime}=\frac{x \cdot g^{\prime}(x)-g(x)}{x^{2}}$

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Video Transcript

it's clear. So when you read here so part, eh? We're gonna apply the product Will were given Why is equal to you, B. That means the derivative of y is equal to you terms the derivative a b plus the derivative of you times be so we get The derivative of y is equal to next from the derivative of g less g of X time the derivative of X which becomes equal to X time the derivative of G bless the dirt plus g of pecs, part B. We have the quotient rule, which is why is equal to X allover g of X. We differentiate using the kosher rule g of x times D over d X minus X times d over the ex Fergie of vets all over G of X square, which is equal to G of X times one minus x times the derivative of G of X All over Do you have back squared which is equal to G of X minus X times the derivative of G of ex all over G F X square. We're on our way. We have part See here we're going to apply the quotient bull as well we get why is equal to G of X over X, so the drive it is is equal to X terms de over de acts g of X minus G of X terms. D over deep X of X over X squared usual, which is equal to X times. The derivative of G minus g affects over X square.

Clarissa N.

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Problem

If $ f $ is a differentiable function, find an ex…

Problem 51 Hard Difficulty

If $ g $ is a differentiable function, find ail expression tor the derivative of each of the following functions.

(a) $ y = xg(x) $

(b) $ y = \frac {x}{g(x)} $

(c) $ y = \frac {g(x)}{x} $

Answer

(a) $y^{\prime}=x g^{\prime}(x)+g(x)$
(b) $y^{\prime}=\frac{g(x)-x \cdot g^{\prime}(x)}{[g(x)]^{2}}$
(c) $y^{\prime}=\frac{x \cdot g^{\prime}(x)-g(x)}{x^{2}}$

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(a) $y^{\prime}=x g^{\prime}(x)+g(x)$(b) $y^{\prime}=\frac{g(x)-x \cdot g^{\prime}(x)}{[g(x)]^{2}}$(c) $y^{\prime}=\frac{x \cdot g^{\prime}(x)-g(x)}{x^{2}}$

Calculus: Early Transcendentals

Heather Z.

Caleb E.

Samuel H.

Joseph L.

Precalculus Review - Intro

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Heather Z.

Caleb E.

Samuel H.

Joseph L.

(a) $y^{\prime}=x g^{\prime}(x)+g(x)$
(b) $y^{\prime}=\frac{g(x)-x \cdot g^{\prime}(x)}{[g(x)]^{2}}$
(c) $y^{\prime}=\frac{x \cdot g^{\prime}(x)-g(x)}{x^{2}}$