Help please??

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Oh this is kind of a fun little exercise :) Chain rule time!
If h(x) = f(f(x)) what is h'(x)=?
I don't know because whats the value of f? Is it 1?

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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?
But here we have a `composition of functions`, so they're testing your understanding of the chain rule.
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`
Oh okay thats right so its f'(f(x)
you took the derivative of the outer function, good. but you didn't multiply by the derivative of the inner function
So I would multiply it by f'(x)?
good good good. so when \(\large\rm h(x)=f(f(x))\) we have found that \(\large\rm h'(x)=f'(f(x))\cdot f'(x)\)
And NOWWWW you can use your chart :)
\[\large\rm h'(\color{orangered}{x})=f'(f(\color{orangered}{x}))\cdot f'(\color{orangered}{x})\]So we want h'(1), yes?\[\large\rm h'(\color{orangered}{1})=f'(f(\color{orangered}{1}))\cdot f'(\color{orangered}{1})\]
And then I would get 18 correct?
Oh I thought we just plug in the values for 1
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?
\[\large\rm h'(1)=f'(\color{red}{f(1)})\cdot f'(1)\]So this is what we're working on first, this red part.
Thats going to be 3 right?
\[\large\rm h'(1)=f'(\color{red}{3})\cdot f'(1)\]Good good.
And then would you multiply it be by 2 since thats the f'(1)
\[\large\rm h'(1)=f'(3)\cdot 2\]2 for f'(1)? Yup sounds good! :) Getting so close!
So then I would have 6 and I would multiply that by 7?
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
F'(3)=7
\[\large\rm h'(1)=7\cdot 2\]Mmmm good! :3
So 14 would be the final answer?
yayyyy good job \c:/
Okay thank you very much!
nppp

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