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BuBbLeS01
Sep30-07, 11:16 AM
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.
EnumaElish
Sep30-07, 11:21 AM
How do you find the derivative of f(x)g(y)?
BuBbLeS01
Sep30-07, 11:24 AM
How do you find the derivative of f(x)g(y)?
use the product rule, f(x)g(y') + g(y)f(x')
rocomath
Sep30-07, 11:39 AM
use the product rule, f(x)g(y') + g(y)f(x')you didn't do the product rule right
BuBbLeS01
Sep30-07, 11:41 AM
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.
EnumaElish
Sep30-07, 12:11 PM
You can apply the chain rule to g(y) (for example, g'(y) dy/dx), but you still have a product to sort through.
BuBbLeS01
Sep30-07, 12:13 PM
so if you use the product rule you would get...
(2sinx -siny') + (2cosxcosy) = 0
EnumaElish
Sep30-07, 12:18 PM
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) = ...
BuBbLeS01
Sep30-07, 12:31 PM
f(x)g(y') + f(x')g(y)
sinxy' + 2cosxcosy
BuBbLeS01
Sep30-07, 01:12 PM
okay thats not right... it should be...
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny
EnumaElish
Sep30-07, 01:23 PM
f(x)g(y') + f(x')g(y)It should be f(x)g'(y)y' + f '(x)g(y).
BuBbLeS01
Sep30-07, 01:26 PM
oh ok...so is that answer right now?
2cosxcosy - 2sinxsinyy' = 0
2cosxcosy - 2y'sinxsiny = 0
y' = -2cosxcosy/-2sinxsiny
EnumaElish
Sep30-07, 01:30 PM
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.
BuBbLeS01
Sep30-07, 01:34 PM
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
EnumaElish
Sep30-07, 01:41 PM
I think you're right. You can cancel out the minuses.
BuBbLeS01
Sep30-07, 01:43 PM
ok thank you!
rocomath
Sep30-07, 06:49 PM
ok thank you!you still have 1 more step

what is cosine/sine?
HallsofIvy
Sep30-07, 07:59 PM
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 sinxcosx]'= 2 [(sin(x)' cos(x)+ sin(x) cos(x)']