## DLS 3 years ago Find dy/dx !

1. DLS

$y=\cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}$

2. DLS

I want to build my logic and way of attacking problems !

3. DLS

$\large y= \cos(ax^{2}+bx+c)+\sin^{3}\sqrt{ax^{2}+bx+c}$

4. anonymous

errr implicit derivative... only with variables? D:

5. anonymous

no actual points? xD its going to get messy

6. DLS

haha,i know..and no sorry :D

7. anonymous

alright i'll try it out, I'm taking cal 1, good review for my test. Hum first, the whole thing is an addition of 2 big terms, so find the derivative of bot terms. The first one is a chain rule : -sin s(ax ^2+bx+c) * (2ax + b)

8. DLS

yes!! i did that

9. DLS

what about the sin^3 part :S

10. anonymous

secondly comes the other term. Because its a root we'll simplify it to: sin ^3 (ax ^2+bx+c)^(1/2) Again the chain rule So we get cos^3(ax ^2+bx+c)^(1/2) * (ax ^2+bx+c)^(1/2)' Therefore we have ANOTHER chain rule: 1/2 (ax ^2+bx+c)^-(1/2) * (2ax+b)

11. anonymous

so you have what I think of a chainception

12. anonymous

Does it look right to you? I don't know if I got it right :) but there is 2 chain rule involved here

13. DLS

Isnt this what we have atm? $\frac{dy}{dx}= -\sin(ax^{2}+bx+c) \times (2ax+b) + \cos^{3} \sqrt{ax^{2}+bx+c} \times \frac{1}{2 \sqrt{ax^{2}+bx+c}} \times 2ax+b$

14. DLS

in the end thats 2ax+b

15. anonymous

yeah thats it. Thats what you should have in my opinion :)

16. anonymous

the chain rule applied to the second term yields $\large \color{blue} 3 \cos ^\color{red} 2 \sqrt{ax^2+bx+c}\frac{ 1 }{ 2\sqrt{ax^2+bx+c} }\left( 2ax+b\right)$

17. DLS

im doubtful about derivative of sin^3 as cos^3

18. anonymous

nope I was right :D hehehe

19. DLS

HA! :D

20. anonymous

Sirm3d. you take the derivative of the exponent of cos too?

21. anonymous

its clearer when you write it as $\large \left[ \sin (ax^2+bx+c)^{1/2} \right]^3$

22. DLS

sugoi!

23. anonymous

ah silly me I was thinking of sin(3x) or something

24. DLS

so u failed in ur test :P

25. anonymous

ugh. let me retype my answer.$3\left[ \sin (ax^2+bx+c)^{1/2} \right]^2\space \cos (ax^2+bx+c)^{1/2}\space \frac{ 1 }{ 2(ax^2+bx+c)^{1/2} }\left(2ax+b\right)$

26. DLS

O___O

27. anonymous

Yeah I was missing a whole term cause of it. Thanks sirm3d for correcting me!

28. DLS

okay!

29. anonymous

that is the derivative of the second term only. the derivative of the first term provided by @MarcLeclair is correct.

30. DLS

why did we have a cos there?

31. DLS

$\large \sin^{3}x$ suppose we have this

32. DLS

wont it be onlly 3cos^2x

33. anonymous

3(sin x)^2 * cos x by chain rule

34. anonymous

you should write it as (sin x)^3, then apply the chain rule

35. DLS

okay..!

36. DLS

thanks!

37. DLS

i get to learn a lot from every ques! :D

38. anonymous

|dw:1354535975146:dw|

39. DLS

yes i got it!