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Explain the relationship between the answers in parts (a) and (b) of Exercises16 and 17. Hint: let $g(x)=1 / f(x)$ and find $g^{\prime}(x)$ using the quotient rule.

$g^{\prime}(x)=\frac{-f^{\prime}(x)}{(f(x))^{2}}$f and g= zeroy=reciprocals.

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 5

Derivative Rules 2

Derivatives

Missouri State University

Campbell University

Boston College

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.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Let $f(x)=x^{2}+3 x,$ and …

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Using the Quotient Rule In…

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(A) Find $f^{\prime}(x)$ u…

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Consider the functions def…

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eso the hitting. This problem tells you to define G events as one over ffx eso. This is sort of your definition of being reciprocal. Okay, so this is I'll just write death. It's a definition of being a reciprocal is to flip flip things. And that's what you were doing in the previous problems. So when I ask you for the derivative, you do the quotient rule which the derivative of the top zero leave the bottom alone with zero times anything is nothing minus yeah. You leave the top alone, you multiply by the derivative of the bottom all over the denominator, which is the FX squared. What you should notice there is that you can simplify that to just be negative f prime of X over FX Square. So how is this going to help us? I'll circle that is because Well, if you look at this problem, the derivative of G equals zero when the derivative of F also equal zero. Um, so even though they're reciprocal, they share the same horizontal tangents. I guess I should say X values, because if the derivative of F equals zero, then that implies the derivative of G equals zero eso If the both the derivatives equals zero, then horizontal tangents are the same when their reciprocal of each other. And hopefully I haven't over. Explain this. But I've said enough to help you get something on paper. Um, there you go. Circle that as well.

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