calyne 3 years ago Find dy/dx by implicit differentiation: e^y cos(x) = 1 + sin(xy)

1. calyne

i got [sin(y) cos(x) + e^y sin(x)] / [e^y cos(x) - sin(x) cos (y)] but the textbook answer is [e^y sin(x) + y cos(xy)] / [e^y cos(x) - x cos(xy)] so am i close or am i there and if those answers are the same how are they equal show me

2. eseidl

$y'e^y \cos x -e^y \sin x=0+\cos(xy)(x)y'+\cos(xy)y$so,$y'=\frac{e^ysinx+\cos(xy)y}{e^ycosx-xcos(xy)}$

3. eseidl

not sure where you are getting siny*cosx terms...they are incorrect

4. calyne

why is d/dx [sin(xy)] equal to cos(xy)(x)y' + cos(xy)y ???? and i know the derivative of cos x, i had it as negative

5. calyne

i thought it was like sin(x)*sin(y)....... so that's how i tried to get the derivative oops but i still don't get what it is then

6. eseidl

using the product rule I will differentiate d/dx [sin(xy)] = d/dx [sin(xy)]*d/dx[xy]

7. eseidl

we get: cos (xy)*y

8. calyne

no ok i know how

9. calyne

but sin isn't its own fluttering thing

10. calyne

oh is it

11. calyne

oh wait flutter i'm reallll rusty on my trig i never learned it properly professor never got around to it so i had to cram myself for the departmental exam

12. calyne

so yeah thanks anyway

13. calyne

idk

14. eseidl

this is the first term. but we have to do it again with respect to y and apply to chain rule so, the second term is:$d/dx[\sin(xy)]=\cos(xy)*d/dy(xy)*dy/dx$

15. eseidl

looks messy but, you get:$xcos(xy)*y'$

16. eseidl

now you sum them (this is just the usual product rule).

17. eseidl

so really we are applying the product rule AND the chain rule to get this

18. calyne

wait WHAT the flutter d/dx (xy) is x*d/dx(y)(* dy/dx)+y*d/dx(x) so it's x(1) dy/dx + y(1) soooo the answer would be xcos(xy) dy/dx + ycos(xy)

19. eseidl

as an easier example of this process consider differentiating xy implicitly. d/dx[xy]=y+(d/dy)(dy/dx)(xy) =y+xy'

20. calyne

why

21. eseidl

maybe it's easier if we write xy=xf(x) where y= f(x) now differentiate it: d/dx[x*f(x)]=f(x)+xf'(x) or, =y+xy'

22. eseidl

this is just the product rule, right? when we implicitly differentiate we assume that y is a function of x. y=f(x)

23. calyne

yeah yeah but flutter I DON"T EVEN KNOW THE flutterING IMPLICIT DIFFERENTIATION RULES MY TEXTBOOK SUCKS AND MY PROFESSOR SUCKS AND THERE ARE NO flutterING RULES IN THE TEXTBOOK FOR SOME REASON IT JUST SHOWS EXAMPLES WITHOUT EVEN EXPLAINING so ALL I KNOW is that you stick y' to multipy the derivative of any term with a y in it

24. calyne

that's fluttering it i don't know why or how or the details or fluttering anything flutter

25. eseidl

when we have x*f(x) the overall change must take into account the rate of change of x and the rate of change of f(x). the product rule says that the overall rate of change of x*f(x) is the rate of change of x (i.e., =1) time f(x) PLUS x*(the rate of change of f(x)=f'(x))

26. calyne

and wtf so the product rule for sin(xy) how does that work too

27. calyne

product rule is d/dx[uv] = u d/dx[v] + v d/dx (u). wtf is u and what is v first of all.

28. eseidl

remember we work from the outside to the inside. if we has sin(8x), working outside in, we get cos(8x)*d/dx(8x)=8cosx.

29. calyne

in sin(xy) how does that translate

30. calyne

but xy is x*y

31. calyne

ugh

32. eseidl

yeah, the only difference between sin(8x) and sin(xy) is y is a function of x and 8 is not

33. eseidl

so we have to account for this in the overall rate of change. sin(xy)=cos(xy)*(d/dx)(xy)

34. eseidl

now...the crucial thing to realize here is both x and y are functions of x in the above

35. eseidl

so we have cos(xy) times d/dx(xy). we use the product rule on this second term. d/dx(xy)=y+xy' So overall we get cos(xy)[y+xy']=ycos(xy)+xy'cos(xy

36. eseidl

got to go, hope this helps!