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anonymous

  • one year ago

Help please??

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  1. zepdrix
    • one year ago
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    Oh this is kind of a fun little exercise :) Chain rule time!

  2. zepdrix
    • one year ago
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    If h(x) = f(f(x)) what is h'(x)=?

  3. anonymous
    • one year ago
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    I don't know because whats the value of f? Is it 1?

  4. zepdrix
    • one year ago
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    no no don't look at the chart yet :O just apply your chain rule. if y = g(x) then y' = g'(x) yes?

  5. zepdrix
    • one year ago
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    But here we have a `composition of functions`, so they're testing your understanding of the chain rule.

  6. zepdrix
    • one year ago
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    Our chain rule tells us that if \(\large\rm y=f(g(x))\) then \(\large\rm y'=f'(g(x))\cdot g'(x)\). Remember that? You take the derivative of the outer function `f in the example` and multiply by the derivative of the inner function `g in the example`

  7. anonymous
    • one year ago
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    Oh okay thats right so its f'(f(x)

  8. zepdrix
    • one year ago
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    you took the derivative of the outer function, good. but you didn't multiply by the derivative of the inner function

  9. anonymous
    • one year ago
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    So I would multiply it by f'(x)?

  10. zepdrix
    • one year ago
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    good good good. so when \(\large\rm h(x)=f(f(x))\) we have found that \(\large\rm h'(x)=f'(f(x))\cdot f'(x)\)

  11. zepdrix
    • one year ago
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    And NOWWWW you can use your chart :)

  12. zepdrix
    • one year ago
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    \[\large\rm h'(\color{orangered}{x})=f'(f(\color{orangered}{x}))\cdot f'(\color{orangered}{x})\]So we want h'(1), yes?\[\large\rm h'(\color{orangered}{1})=f'(f(\color{orangered}{1}))\cdot f'(\color{orangered}{1})\]

  13. anonymous
    • one year ago
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    And then I would get 18 correct?

  14. anonymous
    • one year ago
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    Oh I thought we just plug in the values for 1

  15. zepdrix
    • one year ago
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    It's going to be kind of tricky following the chart here! Let's start from the inner most function. See that f(1) on the inside? What does that value give us?

  16. zepdrix
    • one year ago
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    \[\large\rm h'(1)=f'(\color{red}{f(1)})\cdot f'(1)\]So this is what we're working on first, this red part.

  17. anonymous
    • one year ago
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    Thats going to be 3 right?

  18. zepdrix
    • one year ago
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    \[\large\rm h'(1)=f'(\color{red}{3})\cdot f'(1)\]Good good.

  19. anonymous
    • one year ago
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    And then would you multiply it be by 2 since thats the f'(1)

  20. zepdrix
    • one year ago
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    \[\large\rm h'(1)=f'(3)\cdot 2\]2 for f'(1)? Yup sounds good! :) Getting so close!

  21. anonymous
    • one year ago
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    So then I would have 6 and I would multiply that by 7?

  22. zepdrix
    • one year ago
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    No no no silly. The 3 can't multiply the 2. The 3 is the argument of the function, he can't be messed with. Look up f'(3) on your chart. No multiply just yet! :O

  23. anonymous
    • one year ago
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    F'(3)=7

  24. zepdrix
    • one year ago
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    \[\large\rm h'(1)=7\cdot 2\]Mmmm good! :3

  25. anonymous
    • one year ago
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    So 14 would be the final answer?

  26. zepdrix
    • one year ago
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    yayyyy good job \c:/

  27. anonymous
    • one year ago
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    Okay thank you very much!

  28. zepdrix
    • one year ago
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    nppp

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