Solving for y prime

1. Sep 30, 2007

BuBbLeS01

1. The problem statement, all variables and given/known data
find the derivative of...
2 sinxcosy = 1

3. The attempt at a solution
(2 cosxcosy) * (cosy')
cos y' = -2 cosxcosy
y' = (-2 cosxcosy)/(cos)

I know thats not right but I am not sure where I am making the mistake.

2. Sep 30, 2007

EnumaElish

How do you find the derivative of f(x)g(y)?

3. Sep 30, 2007

BuBbLeS01

use the product rule, f(x)g(y') + g(y)f(x')

4. Sep 30, 2007

rocomath

you didn't do the product rule right

5. Sep 30, 2007

BuBbLeS01

Can you use the chain rule? i think thats what I was trying to do. I was taking the derivative of the outside then the derivative of the inside.

Last edited: Sep 30, 2007
6. Sep 30, 2007

EnumaElish

You can apply the chain rule to g(y) (for example, g'(y) dy/dx), but you still have a product to sort through.

7. Sep 30, 2007

BuBbLeS01

so if you use the product rule you would get...
(2sinx -siny') + (2cosxcosy) = 0

8. Sep 30, 2007

EnumaElish

Write out the formula for [f(x)g(y)]'. Then substitute in f(x) = 2 sin x, g(y) = cos y, f '(x) = ... and g'(y) = ...

Last edited: Sep 30, 2007
9. Sep 30, 2007

BuBbLeS01

f(x)g(y') + f(x')g(y)
sinxy' + 2cosxcosy

10. Sep 30, 2007

BuBbLeS01

okay thats not right... it should be...
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny

11. Sep 30, 2007

EnumaElish

It should be f(x)g'(y)y' + f '(x)g(y).

12. Sep 30, 2007

BuBbLeS01

oh ok...so is that answer right now?
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny

13. Sep 30, 2007

EnumaElish

Do you think it is right?

Write out f(x)g'(y)y' + f '(x)g(y).

State f(x) and g(x).

State f '(x) and g'(y).

Make the substitutions.

14. Sep 30, 2007

BuBbLeS01

yes I do think its right...

2sinxcosy
f(x) = 2sinx
g(x) = cosy
f'(x) = 2cosx
g'(y) = -sinyy'

2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny

15. Sep 30, 2007

EnumaElish

I think you're right. You can cancel out the minuses.

16. Sep 30, 2007

BuBbLeS01

ok thank you!

17. Sep 30, 2007

rocomath

you still have 1 more step

what is cosine/sine?

18. Sep 30, 2007

HallsofIvy

Staff Emeritus
[2 sinxcosx]'= 2 [(sin(x)' cos(x)+ sin(x) cos(x)']