## anonymous one year ago Help please??

1. zepdrix

Oh this is kind of a fun little exercise :) Chain rule time!

2. zepdrix

If h(x) = f(f(x)) what is h'(x)=?

3. anonymous

I don't know because whats the value of f? Is it 1?

4. 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?

5. zepdrix

But here we have a composition of functions, so they're testing your understanding of the chain rule.

6. 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

7. anonymous

Oh okay thats right so its f'(f(x)

8. zepdrix

you took the derivative of the outer function, good. but you didn't multiply by the derivative of the inner function

9. anonymous

So I would multiply it by f'(x)?

10. zepdrix

good good good. so when $$\large\rm h(x)=f(f(x))$$ we have found that $$\large\rm h'(x)=f'(f(x))\cdot f'(x)$$

11. zepdrix

And NOWWWW you can use your chart :)

12. zepdrix

$\large\rm h'(\color{orangered}{x})=f'(f(\color{orangered}{x}))\cdot f'(\color{orangered}{x})$So we want h'(1), yes?$\large\rm h'(\color{orangered}{1})=f'(f(\color{orangered}{1}))\cdot f'(\color{orangered}{1})$

13. anonymous

And then I would get 18 correct?

14. anonymous

Oh I thought we just plug in the values for 1

15. 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?

16. 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.

17. anonymous

Thats going to be 3 right?

18. zepdrix

$\large\rm h'(1)=f'(\color{red}{3})\cdot f'(1)$Good good.

19. anonymous

And then would you multiply it be by 2 since thats the f'(1)

20. zepdrix

$\large\rm h'(1)=f'(3)\cdot 2$2 for f'(1)? Yup sounds good! :) Getting so close!

21. anonymous

So then I would have 6 and I would multiply that by 7?

22. 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

23. anonymous

F'(3)=7

24. zepdrix

$\large\rm h'(1)=7\cdot 2$Mmmm good! :3

25. anonymous

So 14 would be the final answer?

26. zepdrix

yayyyy good job \c:/

27. anonymous

Okay thank you very much!

28. zepdrix

nppp