anonymous
  • anonymous
Calculus - calculate the derivative from the given information: F(u)=u^3 and g(x)=u= (x+4)/(x-2), find (f o g) ' (3) The answer is -882, but I cannot figure out how to arrive at this value.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Empty
  • Empty
Give it your best shot and I'll help you out since you likely have most of the steps right already.
anonymous
  • anonymous
3(u)^2 * x-2 * 1-x+4*1 ?
anonymous
  • anonymous
f ' u=3u^2 f ' g(x)= 3[(x+4)/(x-2)]^2 g ' (x)= 1/1 ??

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More answers

Jhannybean
  • Jhannybean
\[f(g(x)) = f\left(\frac{x+4}{x+2}\right)\] and \((f(g(3)))' = f'(g(3)) \cdot g'(3)\)
anonymous
  • anonymous
I'm lost
Jhannybean
  • Jhannybean
baby steps. Check it out.
Jhannybean
  • Jhannybean
You're trying to find \((f(g(x)))'\) when \(x=3\) , therefore you have to apply the chain rule. \[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]
Jhannybean
  • Jhannybean
So whats the easiest portion to solve for? that would be \(g(3)\) and \(g'(x)\)
anonymous
  • anonymous
I'm not seeing it It would be helpful if you could brake it down more for me
Jhannybean
  • Jhannybean
\[g(3) = \frac{3+4}{3-2} = \frac{7}{1} = 7\]\[\begin{align} g(x) = (x+4)(x-2)^{-1} &\implies g'(x) = 1(x-2)^{-1} -(x-2)^{-2}(1)(x+4) \\&\implies g'(x) = -\frac{6}{(x-2)^2} \\ &\implies g'(3) =-\frac{6}{(\color{red}{3}-2)^2} \\&\implies g'(3) = -6 \end{align}\]
Jhannybean
  • Jhannybean
Now im a little confused with the whole f(u)=u\(^3\) part.... \(x=u\)?
zepdrix
  • zepdrix
\[\large\rm f(\color{orangered}{u})=(\color{orangered}{u})^3\]And \(\large\rm \color{orangered}{g(x)=u=\frac{x+4}{x-2}}\) So then,\[\large\rm f(\color{orangered}{u})=f(\color{orangered}{g(x)})=\left(\color{orangered}{\frac{x+4}{x-2}}\right)^3\]Ya I think maybe you forgot to chain rule little panda?
Jhannybean
  • Jhannybean
Oh thats what it meant...ok
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}f(g(x))=3\left(\frac{x+4}{x-2}\right)^2\color{royalblue}{\frac{d}{dx}\left(\frac{x+4}{x-2}\right)}\]
zepdrix
  • zepdrix
`g ' (x)= 1/1 ??` No. I guess that's where you ran into trouble. Apply quotient rule or something similar. :)
zepdrix
  • zepdrix
Aw he ran off, prolly had to get some more bamboo
Jhannybean
  • Jhannybean
lol
IrishBoy123
  • IrishBoy123
.
Loser66
  • Loser66
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Loser66
  • Loser66
|dw:1444439482315:dw|

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