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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} $

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00:46

Frank Lin

Calculus 1 / AB

Chapter 3

Differentiation Rules

Section 2

The Product and Quotient Rules

Derivatives

Differentiation

Missouri State University

Oregon State University

University of Nottingham

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

44:57

In mathematics, a differentiation rule is a rule for computing the derivative of a function in one variable. Many differentiation rules can be expressed as a product rule.

02:26

If $g$ is a differentiable…

05:47

03:47

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07:24

If $f$ is a differentiable…

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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