Question

derivative question, please help!

Answer 1

suppose that \[(g(x))^2+6x=x^2g(x)-39\] and that g(4)=7. Find g'(4).

Answer 2

I thought I knew how to get the answer, but my answer wasn't right....

Answer 3

differentiate with respect to x.

since g(x) is a function of x, use the chain rule ...
eg. :

\[\frac{ d }{ dx } g(x) = g '(x) \]
\[\frac{ d }{ dx } (g(x))^{3} = 3(g(x))^{2} g '(x) \]

Answer 4

if there's a product and there is...) use the product rule,
eg.:
\[\frac{ d }{ dx } (g(x) *x^3)= g(x)*3x^2 + x^3* g '(x) \]

Answer 5

Yeah, so then \[\frac{ 1 }{ 2\sqrt{g(4)} }g'(4) +6=[x^2g'(4)+g(4)2x] \right?\]

Answer 6

crap, hold on a sec

Answer 7

:)

Answer 8

equation editor is a bit buggy for me too :)

Answer 9

\[\frac{ 1 }{ 2\sqrt{g(4)} }g'(4)+6=[x^2g'(4)+2xg(4)]\]

Answer 10

right?

Answer 11

hmm

Answer 12

and then you plug in 7 for wherever there's a g(4) and then solve for the variable g'(4)?

Answer 13

is it supposed to be (g(x))^(1/2) in the original?

Answer 14

it says (g(x))^2...

Answer 15

Darn it..... yeah... hold on a sec

Answer 16

\[2\sqrt{g(4)} g'(4)+6=[x^2g'(4)+2xg(4)]\] does that look better?

Answer 17

er no, wait

Answer 18

\[ 2g(4)g'(4)+6=[x^2g'(4)+2xg(4)]\]

Answer 19

good, now since g(4) = 7 , x=4 and g(4) = 7 .... sub.s in and solve for g'(4)

Answer 20

alright, just a minute

Answer 21

So should I have \[y'=\frac{ 14x-6 }{ 7-x^2 }\]

Answer 22

actually wait, I think that's wrong hold on a sec

Answer 23

14-x^2 in the denom.

Answer 24

and x=4

Answer 25

Yep, I just caught that. Thanks so much! :)

Answer 26

sure:)