## anonymous one year ago Calculate the derivative of the given function f(x)=8sin^3(√x) a) f′(x)=24sin^2(√x)cos(√x) b) f′(x)=48sin^2(√x)cos(√x)/x√x c) f′(x)=24cos(√x) d) f′(x)=12sin^2(√x)cos(√x)/√x e) f′(x)=12cos(√x)/√x

1. anonymous

yes, one second

2. anonymous

Sorry, I made a typo. $f(x) = 8\sin^3(x^{1/2})$

3. anonymous

actually could you help me with this really quickly? its more pressing: Evaluate (g∘f)′(6), given that f(4)=4, f(5)=6, f(6)=6, g(4)=5, g(5)=5, g(6)=4, f′(4)=6, f′(5)=5, f′(6)=4, g′(4)=5, g′(5)=5, g′(6)=4 a) 16 b) 19 c) 18 d) 17 e) 15

4. anonymous

sorry, thanks for all your help, im just on a time crunch.

5. anonymous

You're not taking a test are you?

6. anonymous

no, I just have to be somewhere soon

7. anonymous

Oh, alright.

8. anonymous

And this homework is due tomorrow

9. anonymous

So i assume : $(g~\circ ~ f)'(x) \qquad \implies \qquad \frac{d}{dx}(g(f(x)) = g'(f(x)) \cdot f'(x)$

10. anonymous

where $$x=6$$

11. anonymous

Ack, I've got to head off for a little while, I will be back very soon.

12. anonymous

@DanJS @jim_thompson5910 mind taking over? Thanks!

13. DanJS

i am late, are you still working on these things?

14. DanJS

yes, that one would be just the chain rule, and they list the values you may need for the functions and their derivatives $\qquad \frac{d}{dx}(g(f(6)) = g'(f(6)) \cdot f'(6)$

15. DanJS

f(6) = 6 f '(6) = 4 g'(6)=4 so $\qquad \frac{d}{dx}(g(f(x)) = g'(f(x)) \cdot f'(x) = g'(6)*f'(6) = 4*4$

16. DanJS

To find f'(x) you use the chain rule... $\frac{ d }{ dx }f(x) = \frac{ d }{ du }*\frac{ du }{ dx } f(x)$

17. DanJS

If you have a function inside a function , like f(g(x)) f(x) = sin(x^3) I usually think something like; " f ' (x) is equal to derivative of sin(inside function left alone) times the derivative of the inside function" f ' (x) = cos(x^3) * 2x^2 thats it, goodluck

18. DanJS

similar with the given function f(x) = 8*sin(x^(1/2))^3 start with outside and leave inside function same, except here you have to use the rule two times, u^3, sin(u) and x^(1/2) start from outside power function $f ' (x) = 3*8*[\sin(x^{1/2})]^2~*~\cos(x^{1/2})*\frac{ 1 }{ 2 }*x^{-1/2}$

19. DanJS

simplifies to D i think ... check it out,

20. DanJS

keep leaving the inner functions alone, and work outside to inside, power rule - derivative of sin() - derivative of root x

21. DanJS

$[f(g(h(x)))] ' = f'(g(h(x))) *g'(h(x)) * h'(x)$

22. DanJS

you can have as many nested functions as you want, same idea