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If $ f $ and $ g $ are the functions whose graphs are shown, let $ u(x) = f(x) g(x) $ and $ v(x) = f(x)/g(x). $(a) Find $ u'(1). $(b) Find $ v'(5). $
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Calculus 1 / AB
The Product and Quotient Rules
Missouri State University
University of Michigan - Ann Arbor
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.
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.
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Hey, it's clear. So in me rain here. So according to a graph, we could see that F of X is equal to negative one. Have X if X is less than or equal to zero two of X if X is between zero and two included and negative to fifth X minus two plus four. If X is bigger than two than me, ask the derivative, which is equal to negative 1/2 if X is less than zero too, if access between X zero and two and negative to fifth if X is bigger than two and we have G of X, which is equal to to minus X. If X is less than or equal to two than 3/5 times X minus two. If X is bigger than two, we have the derivative of G of X this equal to one negative one. If X is less than two and 3/5 if X is bigger than two. So for party we're gonna find for the derivative of you of one which is equal to the derivative of F of one times the derivative of G of one bless the derivative of up of one times the derivative of G of one. We plug this in to get two times one plus two times negative one. This becomes equal to zero for part B. We're finding for the derivative of V of five. Here we have G of five times the derivative of five minus F five times too derivative of G of five, all over G of five square. You put this in you get 9/5 times and negative to fifth, minus 14 over five times 3 15 a little over 9/5 square, which is equal to negative 20 over 27.
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