Question

Calculus - calculate the derivative from the given information: F(u)=u^3 and g(x)=u= (x+4)/(x-2), find (f o g) ' (3)

The answer is -882, but I cannot figure out how to arrive at this value.

Answer 1

Give it your best shot and I'll help you out since you likely have most of the steps right already.

Answer 2

3(u)^2 * x-2 * 1-x+4*1 ?

Answer 3

f ' u=3u^2
f ' g(x)= 3[(x+4)/(x-2)]^2
g ' (x)= 1/1 ??

Answer 4

\[f(g(x)) = f\left(\frac{x+4}{x+2}\right)\] and \((f(g(3)))' = f'(g(3)) \cdot g'(3)\)

Answer 5

I'm lost

Answer 6

baby steps. Check it out.

Answer 7

You're trying to find \((f(g(x)))'\) when \(x=3\) , therefore you have to apply the chain rule. \[(f(g(x)))' = f'(g(x)) \cdot g'(x)\]

Answer 8

So whats the easiest portion to solve for? that would be \(g(3)\) and \(g'(x)\)

Answer 9

I'm not seeing it
It would be helpful if you could brake it down more for me

Answer 10

\[g(3) = \frac{3+4}{3-2} = \frac{7}{1} = 7\]\[\begin{align} g(x) = (x+4)(x-2)^{-1} &\implies g'(x) = 1(x-2)^{-1} -(x-2)^{-2}(1)(x+4) \\&\implies g'(x) = -\frac{6}{(x-2)^2} \\ &\implies g'(3) =-\frac{6}{(\color{red}{3}-2)^2} \\&\implies g'(3) = -6 \end{align}\]

Answer 11

Now im a little confused with the whole f(u)=u\(^3\) part.... \(x=u\)?

Answer 12

\[\large\rm f(\color{orangered}{u})=(\color{orangered}{u})^3\]And \(\large\rm \color{orangered}{g(x)=u=\frac{x+4}{x-2}}\)
So then,\[\large\rm f(\color{orangered}{u})=f(\color{orangered}{g(x)})=\left(\color{orangered}{\frac{x+4}{x-2}}\right)^3\]Ya I think maybe you forgot to chain rule little panda?

Answer 13

Oh thats what it meant...ok

Answer 14

\[\large\rm \frac{d}{dx}f(g(x))=3\left(\frac{x+4}{x-2}\right)^2\color{royalblue}{\frac{d}{dx}\left(\frac{x+4}{x-2}\right)}\]

Answer 15

`g ' (x)= 1/1 ??`
No.

I guess that's where you ran into trouble.
Apply quotient rule or something similar. :)

Answer 16

Aw he ran off,
prolly had to get some more bamboo

Answer 17

lol

Answer 18

.