## anonymous 4 years ago Let y =(root5(x^8+5)). Find values A and B so that (dy)/(dx) = A(x^8+5)^B(C x^D)

1. myininaya

$y=\sqrt[5]{x^8+5}$ $y=(x^8+5)^\frac{1}{5}$ Use the chain rule to find y'

2. anonymous

what if I need to find A,B,C, and D? what do I plug in?

3. myininaya

You need to find y' first to find A,B,C,and D

4. anonymous

I havent learned the chain rule yet

5. myininaya

$[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule }$

6. myininaya

Well you need the chain rule to do this problem.

7. myininaya

what is g(x) and what is f(x) if $f(g(x))=(x^8+5)^\frac{1}{5}$

8. anonymous

g(x)= 8x f(x)= 5^1/5

9. myininaya

No g(x) is the inside function and f(x) is the outside function

10. anonymous

oh ok g(x)= 8x^7

11. myininaya

$g(x)=x^8+5 ; f(x)=x^\frac{1}{5}$ $g'(x)=8x^7+0 ; f(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}$

12. anonymous

so I plug this in in the equation now

13. myininaya

this one $y'=[f(g(x))]'=f'(g(x)) \cdot g'(x) \text{ chain rule }$ yes

14. anonymous

1/5x^-4/5

15. myininaya

$y'=f'(x^8+5) \cdot 8x^7$

16. anonymous

now just multiply, right?

17. myininaya

you need to plug x^8+5 into f'

18. anonymous

oh ok

19. myininaya

$g'(x)=8x^7+0 ; f'(x)=\frac{1}{5}x^{\frac{1}{5}-1}=\frac{1}{5}x^\frac{-4}{5}$

20. myininaya

f' is that second one I wrote down just now

21. myininaya

$y'=f'(x^8+5) \cdot 8x^7=\frac{1}{5}(x^8+5)^\frac{-4}{5} \cdot 8x^7$

22. myininaya

now you should be able to put it in the form above to get your A,B,C, and D

23. anonymous

i'll try it

24. myininaya

actually it is already in that form

25. anonymous

so just look for the A,B,C,D then right?

26. myininaya

yes