anonymous
  • anonymous
Help please??
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
Oh this is kind of a fun little exercise :) Chain rule time!
zepdrix
  • zepdrix
If h(x) = f(f(x)) what is h'(x)=?
anonymous
  • anonymous
I don't know because whats the value of f? Is it 1?

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zepdrix
  • zepdrix
no no don't look at the chart yet :O just apply your chain rule. if y = g(x) then y' = g'(x) yes?
zepdrix
  • zepdrix
But here we have a `composition of functions`, so they're testing your understanding of the chain rule.
zepdrix
  • zepdrix
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`
anonymous
  • anonymous
Oh okay thats right so its f'(f(x)
zepdrix
  • zepdrix
you took the derivative of the outer function, good. but you didn't multiply by the derivative of the inner function
anonymous
  • anonymous
So I would multiply it by f'(x)?
zepdrix
  • zepdrix
good good good. so when \(\large\rm h(x)=f(f(x))\) we have found that \(\large\rm h'(x)=f'(f(x))\cdot f'(x)\)
zepdrix
  • zepdrix
And NOWWWW you can use your chart :)
zepdrix
  • zepdrix
\[\large\rm h'(\color{orangered}{x})=f'(f(\color{orangered}{x}))\cdot f'(\color{orangered}{x})\]So we want h'(1), yes?\[\large\rm h'(\color{orangered}{1})=f'(f(\color{orangered}{1}))\cdot f'(\color{orangered}{1})\]
anonymous
  • anonymous
And then I would get 18 correct?
anonymous
  • anonymous
Oh I thought we just plug in the values for 1
zepdrix
  • zepdrix
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?
zepdrix
  • zepdrix
\[\large\rm h'(1)=f'(\color{red}{f(1)})\cdot f'(1)\]So this is what we're working on first, this red part.
anonymous
  • anonymous
Thats going to be 3 right?
zepdrix
  • zepdrix
\[\large\rm h'(1)=f'(\color{red}{3})\cdot f'(1)\]Good good.
anonymous
  • anonymous
And then would you multiply it be by 2 since thats the f'(1)
zepdrix
  • zepdrix
\[\large\rm h'(1)=f'(3)\cdot 2\]2 for f'(1)? Yup sounds good! :) Getting so close!
anonymous
  • anonymous
So then I would have 6 and I would multiply that by 7?
zepdrix
  • zepdrix
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
anonymous
  • anonymous
F'(3)=7
zepdrix
  • zepdrix
\[\large\rm h'(1)=7\cdot 2\]Mmmm good! :3
anonymous
  • anonymous
So 14 would be the final answer?
zepdrix
  • zepdrix
yayyyy good job \c:/
anonymous
  • anonymous
Okay thank you very much!
zepdrix
  • zepdrix
nppp

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