## anonymous one year ago 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.

1. Empty

2. anonymous

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

3. anonymous

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

4. Jhannybean

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

5. anonymous

I'm lost

6. Jhannybean

baby steps. Check it out.

7. Jhannybean

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

8. Jhannybean

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

9. anonymous

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

10. Jhannybean

$g(3) = \frac{3+4}{3-2} = \frac{7}{1} = 7$\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}

11. Jhannybean

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

12. zepdrix

$\large\rm f(\color{orangered}{u})=(\color{orangered}{u})^3$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?

13. Jhannybean

Oh thats what it meant...ok

14. zepdrix

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

15. zepdrix

g ' (x)= 1/1 ?? No. I guess that's where you ran into trouble. Apply quotient rule or something similar. :)

16. zepdrix

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

17. Jhannybean

lol

18. IrishBoy123

.

19. Loser66

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20. Loser66

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