## SandyHearts Group Title Find the derivative: sin(xcosx) so the answer is after being simplified (cosx-xsinx)cos(xcosx) I have no idea. how. 2 years ago 2 years ago

1. Outkast3r09 Group Title

this is chain rule

2. Outkast3r09 Group Title

$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ where your y=sin(u) and your u=xcos(x)

3. SandyHearts Group Title

ok so: sin(xcosx)(-sinx)(1)+cos(xcosx)

4. SandyHearts Group Title

why is that not the answer I am getting :(

5. Outkast3r09 Group Title

alright so derivative of sin(u) = cos(u) so $\frac{dy}{du}=cos(u)$ next do the product rule with your u=xcos(x) u'=vw'+wv'=-xsin(x)+cos(x)

6. Outkast3r09 Group Title

so you end up with cos(xcos(x)(-xsin(x)+cos(x))

7. Outkast3r09 Group Title

which is the same as (cos(x)-xsin(x))cos(xcos(x))

8. Outkast3r09 Group Title

does this make sense?

hi |dw:1340953644460:dw|

10. SandyHearts Group Title

from this cos(xcos(x)(-xsin(x)+cos(x)) all i can get is cos(xcos(x)(-xsin(x)) where does the +cos(x) come from

11. myininaya Group Title

$\cos( x \cos(x)) \cdot (x \cos(x))' \text{ by chain rule }$

12. myininaya Group Title

Now use product rule for that one part

13. SandyHearts Group Title

OHHHHHHHH

14. SandyHearts Group Title

>.<

15. SandyHearts Group Title

I see @Outkast3r09 tried explaining that to me but I was really not understanding all the notations you used. Thank you all so much.

16. SandyHearts Group Title

Sorry @Outkast3r09 I didn't understand the first time you explained that to me.

17. Outkast3r09 Group Title

it comes from the chain rule do you understand this $\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ if you name your inner function u and your outer function y, your outer function will become y=f(u).. so now you find the derivative of your y in respects to u

18. Outkast3r09 Group Title

then you take your u which is u=f(x)... and find the derivative in respects to x.. then multply them together

19. Outkast3r09 Group Title

since your u=xcos(x) you have to use product rule to find the derivative

20. SandyHearts Group Title

I'm going to try to work it out again

21. SandyHearts Group Title

y=sin(xcosx) y'= cos(xcosx)(x)(-sinx)+cosx That's the same thing right? I think I got it.

22. SandyHearts Group Title

idek how i got that anymore.

23. Outkast3r09 Group Title

hehe alright start by turning your outer function into y=f(u)

24. Outkast3r09 Group Title

by letting your u be the innre function

25. dpaInc Group Title

ok... let's start from the beginning.. $\huge y=sin(xcosx)$ $\huge y'=cos(xcosx)\cdot [xcosx]'$ by the chain rule...

26. Outkast3r09 Group Title

i'd start by using the actual theorem dpalnc

27. dpaInc Group Title

now... $\huge [xcosx]'=x'\cdot cosx + x \cdot [cosx]'$ $\huge [xcosx]'=1\cdot cosx + x \cdot (-sinx)$ $\huge [xcosx]'= cosx - x sinx$ by the product rul...

28. dpaInc Group Title

now putting these together...

29. dpaInc Group Title

$\huge y'=cos(xcosx)\cdot (cosx - xsinx)$

30. SandyHearts Group Title

THANK YOU!!!!!

31. SandyHearts Group Title

I got it. 100%

32. dpaInc Group Title

yw...:)

33. dpaInc Group Title

just take it one step at a time and you should be good...:)

34. SandyHearts Group Title

:) Alright I will.

35. dpaInc Group Title

:)