Find the derivative of f(x)=-9/x at x=-4 @agent0smith @jim_thompson5910 @Phebe @e.mccormick
First find the derivative of f(x)=-9/x
That's excatly what i dont know how to do >.<
think of -9/x as -9x^(-1)
then use the idea that if y = x^n, then y ' = n*x^(n-1)
That is called the power rule. Have you been taught that yet?
i have not
OK. What are you using at this point to do derivatives? The limit definition?
i think i got is ...4/9?
I do not get that answer. Are you using \(f'(x)=\lim\limits_{\Delta x\rightarrow 0}\dfrac{f(x+\Delta x)-f(x)}{\Delta x}\) to find derivatives?
yes.. im redoing it
16/9
OK.
i think im wrong again
How did you get it with 16 on the top and 9 on the bottom?
i honestly think im doing the work wrong
i redid it... 9/16?
please tell me thats right because im literally dying lolol
You got it. I am going to show you how the thing Jim talked about works: \(\dfrac{d}{dx}x^n=nx^{n-1}\) \(\dfrac{d}{dx}\dfrac{-9}{x} \implies \dfrac{d}{dx}-9x^{-1}\) \(\dfrac{d}{dx}-9x^{-1} \implies -9\left( \dfrac{d}{dx}x^{-1}\right )\) Now I use the rule above. \(-9\left( \dfrac{d}{dx}x^{-1}\right) = -9\left( -1x^{-1-1}\right)\) Which simplifies to: \(9x^{-2}\) or: \(\dfrac{9}{x^2}\)
And at -4, that is \(\dfrac{9}{16}\)
Wow that is way easier than what i was trying to do.. THANK YOU SO MUCH♥
Well, you need to learn why it works still. They teach math in steps. You learn addition before multiplication. Multiplication is just repeated addition. You will learn the power rule soon enough. But seeing it, you have a way to check your work.
YES honestly thank you, math is my weakness haha
@e.mccormick
Find the derivative of f(x) = 4/ x at x = 2 I GOT 4 can you tell me if thats good?
@jim_thompson5910 @e.mccormick
deriving 4/x gives you ???
think of 4/x as 4x^(-1)
Wait... im i completly wrong?
what did you get for f ' (x)
-1
when you use the power rule to derive x^(-1), you get ???
wait... -4
Here's how I think of the power rule |dw:1400110066022:dw|
pull down the exponent |dw:1400110207950:dw|
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