## anonymous 5 years ago Find y' assuming that the equation determines a differentiable function f such that y= f(x) sin^2 3y=x+y-1

1. amistre64

$$sin^2(3y)$$ ??

2. anonymous

its implicit differentiation

3. amistre64

yeah; but whats the equation?

4. anonymous

yes amistre that is correct

5. anonymous

its kind of assumed its sin^2 (3y) lol

6. anonymous

its unlikely to be sin^2(3) y

7. amistre64

$3y' 2\cos(3y)=1+y'$ $6y'\cos(3y)-y'=1$ $y'(6\cos(3y)-1)=1$ $y' = \frac{1}{6\cos(3y)}$

8. amistre64

thats wrong ....

9. anonymous

?

10. amistre64

i did sin^2(3y) as tho it was simply sin(3y)

11. anonymous

ohh yeh, I see

12. anonymous

alot of chain rules in that question

13. anonymous

its only once elecengineer

14. anonymous

lol

15. amistre64

$3y'.2(\sin(3y)).cos(3y)=1+y'$ $y'.6\sin(3y)\cos(3y)-y'= 1$ $y'(6\sin(3y)\cos(3y)-1)= 1$ $y' = \frac{1}{6\sin(3y)\cos(3y)-1}$ maybe

16. amistre64

3 times actually

17. amistre64

$u^2.sin(v).3y$

18. anonymous

also, you couls apply double angle formula for sin to simplify the bottom :P

19. amistre64

yup

20. anonymous

im trying to figure out what you did amistre

21. anonymous

consider the LHS first, the derivative of the RHS is easy as

22. anonymous

so ( sin(3y) )^2 apply the chain rule , bring the power down in front , leave the inside of the bracket alone , reduce the power by one , then multiplt by the derivative of the inside

23. anonymous

so d/dx [ (sin(3y) ) ^2 ] = 2 sin(3y) [d/dx ( sin(3y) ) ]

24. anonymous

= 2 sin(3y) 3y' cos(3y)

25. amistre64

say you wanna derive: u^2; thats simple right? but: u = sin(v) and v = 3y

26. anonymous

when you differentiate sin(3y) you take the derivative of the inside with respect to x, and then change the sin function to a cos function

27. anonymous

the derivative of 3y with respect to x is 3 (dy/dx) = 3y'

28. anonymous

i get the 2(sin3y)(cos y) but not the 3y'

29. amistre64

${du\over dx}={du\over dv}{dv\over dy}{dy\over dx}$ $\frac{d(sin^2(3y))}{dx}=2sin(3y)*cos(3y)*3y'$

30. amistre64

you dont understand why we keep something hta tyou were taught to throw out

31. anonymous

:S?

32. amistre64

$y = 2x^3 \rightarrow y' = 6x^2 x'$

33. anonymous

oh...

34. amistre64

what did the $$y$$ derive to? $$y'$$ right?

35. amistre64

3y derives to: $$3 y'$$ $$5y^2$$ derives to $$10y .y'$$

36. anonymous

so the derivative of sin 3y is 3 sin 3y then 3y' cos 3y?

37. amistre64

you are used to throwing out the x' bit; but that is only because $$dx\over dx$$=1

38. anonymous

i know that ist the trig function thats throwing me off

39. amistre64

sin(3y) derives to 3y' cos(3y)

40. anonymous

yes...thats what confused me or should i say confuses

41. amistre64

the sin to cos? or the innards to the outside?

42. anonymous

the innards the 3y inside the sin 3y

43. amistre64

the innards have a controling part in the function and have to be accounted for. They are the driving force that produces the output for the sin function

44. amistre64

the 'chain' rule is just accounting for all the inputs and outputs thru the equation that eah have a controling part

45. anonymous

ok i guess i have to practice some more examples of them

46. amistre64

try it on simple ones first: like, compound functions

47. amistre64

interactmath.com can help; its a free practice math site

48. anonymous

THANKS AGAIN SO MUCH AMISTRE AND ELECENGINEER

49. amistre64

youre welcome :)

50. anonymous

can you help me with the question i posted earlier today