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anonymous

  • one year ago

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.

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  1. Empty
    • one year ago
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    Give it your best shot and I'll help you out since you likely have most of the steps right already.

  2. anonymous
    • one year ago
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    3(u)^2 * x-2 * 1-x+4*1 ?

  3. anonymous
    • one year ago
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    f ' u=3u^2 f ' g(x)= 3[(x+4)/(x-2)]^2 g ' (x)= 1/1 ??

  4. Jhannybean
    • one year ago
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    \[f(g(x)) = f\left(\frac{x+4}{x+2}\right)\] and \((f(g(3)))' = f'(g(3)) \cdot g'(3)\)

  5. anonymous
    • one year ago
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    I'm lost

  6. Jhannybean
    • one year ago
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    baby steps. Check it out.

  7. Jhannybean
    • one year ago
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    You're trying to find \((f(g(x)))'\) when \(x=3\) , therefore you have to apply the chain rule. \[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]

  8. Jhannybean
    • one year ago
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    So whats the easiest portion to solve for? that would be \(g(3)\) and \(g'(x)\)

  9. anonymous
    • one year ago
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    I'm not seeing it It would be helpful if you could brake it down more for me

  10. Jhannybean
    • one year ago
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    \[g(3) = \frac{3+4}{3-2} = \frac{7}{1} = 7\]\[\begin{align} g(x) = (x+4)(x-2)^{-1} &\implies g'(x) = 1(x-2)^{-1} -(x-2)^{-2}(1)(x+4) \\&\implies g'(x) = -\frac{6}{(x-2)^2} \\ &\implies g'(3) =-\frac{6}{(\color{red}{3}-2)^2} \\&\implies g'(3) = -6 \end{align}\]

  11. Jhannybean
    • one year ago
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    Now im a little confused with the whole f(u)=u\(^3\) part.... \(x=u\)?

  12. zepdrix
    • one year ago
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    \[\large\rm f(\color{orangered}{u})=(\color{orangered}{u})^3\]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?

  13. Jhannybean
    • one year ago
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    Oh thats what it meant...ok

  14. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}f(g(x))=3\left(\frac{x+4}{x-2}\right)^2\color{royalblue}{\frac{d}{dx}\left(\frac{x+4}{x-2}\right)}\]

  15. zepdrix
    • one year ago
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    `g ' (x)= 1/1 ??` No. I guess that's where you ran into trouble. Apply quotient rule or something similar. :)

  16. zepdrix
    • one year ago
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    Aw he ran off, prolly had to get some more bamboo

  17. Jhannybean
    • one year ago
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    lol

  18. IrishBoy123
    • one year ago
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    .

  19. Loser66
    • one year ago
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    |dw:1444439467979:dw|

  20. Loser66
    • one year ago
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    |dw:1444439482315:dw|

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