Question

Let y =(root5(x^8+5)).

Find values A and B so that (dy)/(dx) = A(x^8+5)^B(C x^D)

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

I havent learned the chain rule yet

Answer 5

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

Answer 6

Well you need the chain rule to do this problem.

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

oh ok g(x)= 8x^7

Answer 11

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

Answer 12

so I plug this in in the equation now

Answer 13

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

Answer 14

1/5x^-4/5

Answer 15

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

Answer 16

now just multiply, right?

Answer 17

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

Answer 18

oh ok

Answer 19

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

Answer 20

f' is that second one I wrote down just now

Answer 21

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

Answer 22

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

Answer 23

i'll try it

Answer 24

actually it is already in that form

Answer 25

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

Answer 26

yes