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anonymous
 one year ago
Help please??
anonymous
 one year ago
Help please??

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Oh this is kind of a fun little exercise :) Chain rule time!

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2If h(x) = f(f(x)) what is h'(x)=?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't know because whats the value of f? Is it 1?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2no no don't look at the chart yet :O just apply your chain rule. if y = g(x) then y' = g'(x) yes?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2But here we have a `composition of functions`, so they're testing your understanding of the chain rule.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2Our 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`

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh okay thats right so its f'(f(x)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2you took the derivative of the outer function, good. but you didn't multiply by the derivative of the inner function

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I would multiply it by f'(x)?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2good good good. so when \(\large\rm h(x)=f(f(x))\) we have found that \(\large\rm h'(x)=f'(f(x))\cdot f'(x)\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2And NOWWWW you can use your chart :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm h'(\color{orangered}{x})=f'(f(\color{orangered}{x}))\cdot f'(\color{orangered}{x})\]So we want h'(1), yes?\[\large\rm h'(\color{orangered}{1})=f'(f(\color{orangered}{1}))\cdot f'(\color{orangered}{1})\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then I would get 18 correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh I thought we just plug in the values for 1

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2It'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?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm h'(1)=f'(\color{red}{f(1)})\cdot f'(1)\]So this is what we're working on first, this red part.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thats going to be 3 right?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm h'(1)=f'(\color{red}{3})\cdot f'(1)\]Good good.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And then would you multiply it be by 2 since thats the f'(1)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm h'(1)=f'(3)\cdot 2\]2 for f'(1)? Yup sounds good! :) Getting so close!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So then I would have 6 and I would multiply that by 7?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2No 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

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.2\[\large\rm h'(1)=7\cdot 2\]Mmmm good! :3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 14 would be the final answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay thank you very much!
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