Jo L.

asked • 05/14/20

Find the rate of change for f(x) = -3 / (2x+1), when x = -2

Nathan G.

Hey Jo, do you know how to take the derivative yet? Or does this require the long formula?
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05/14/20

Jo L.

Hi, Nathan! Yes, I think! Thanks!
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05/15/20

Jo L.

My bad, I do not know how to take the derivative; how do you do it? Or should I keep using the long formula?
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05/15/20

1 Expert Answer

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Alexandra L. answered • 05/15/20

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In order to answer rate of change questions, you have to take the derivative. To make the problem simpler, factor out the 3 and use exponents.


f(x)= -3 (2x+1)-1


Now you just proceed to take the derivative normally and don't forget to use chain rule! In the end, plug in -2 for x in your f'(x) to get the answer. Let me know if you need more help

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Jo L.

Alexandra, thank you for your help! I just realized that I do not know how to "take a derivative". How do you do that?
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05/15/20

Alexandra L.

tutor
Taking a derivative will change a little depending on what operations are used in an equation. There are "derivative rules" that help you find the derivative quickly. A derivative (f'(x) or y') will tell you the rate of change at that specific point in the graph. Let's start with some of the basic rules. Power Rule: If y=x^n then y'=nx^(n-1). So bring down the exponent in front of your x and subtract 1 from the exponent. In this problem, replace let 2x+1=u to make it easier on yourself. u=2x+1 f(x)=-3u^-1 f'(x)=-3(-1)u^(-1-1) f'(x)=3u^-2 now you have to plug x in again! but before you do that, you have to find the derivative of u. u=2x+1. Apply power rule again for the 2x part and remember that the derivative of a constant (the 1) is 0. So u'=2. When you plug in x to your equation, you have to use chain rule which means you have to also account for the derivative of u. The way you do that is by multiplying it. f'(x)= [3(2x+1)^-2] x 2 f'(x)= 6(2x+1)^-2 This is all pretty hard to explain on here, but I hope it helped. Let me know if you want to schedule a lesson to get a better grasp of derivatives.
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05/15/20

Jo L.

Haha, Alexandra, thank you so much! I understand it well with your help! :)
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05/15/20

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