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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} $

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

00:46

Frank L.

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 2

The Product and Quotient Rules

Derivatives

Differentiation

Oregon State University

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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.

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