Posted by **josephine** on Wednesday, February 29, 2012 at 10:25pm.

find y' by implicit differentiation.

i dont understand, its so complicate.

1. 4cos(x-y)=3ysinx

2. 4sin(x-y) =3ysinx

- Calculus implicit -
**David**, Wednesday, February 29, 2012 at 10:43pm
4sin(x-y)*d(x-y)=3y' + 3ycosx

On the left I applied the chain rule, and on the right i used the product rule.

4sin(x-y)*(1-y')=3y' + 3ycosx

Differentiate x - y and get 1 - y'

Then do algebra to solve for y'

4sin(x-y) - y'*4sin(x-y) = 3y' + 3ycosx

3y' + y'*4sin(x-y) = 4sin(x-2) - 3ycosx

y' = [4sin(x-2)-3ycosx]/[3+4sin(x-y)]

Try the other one yourself.

- Calculus implicit -
**josephine**, Wednesday, February 29, 2012 at 10:58pm
ok i'll try another, thanks

- Calculus implicit -
**Damon**, Wednesday, February 29, 2012 at 11:08pm
oh

d/dx cos u = - sin u du/dx

You forgot the - sign

- Calculus implicit - oops -
**Steve**, Thursday, March 1, 2012 at 10:58am
also missed on the right side :-(

4cos(x-y)=3ysinx

-4sin(x-y)(1-y') = 3sinx*y' + 3ycosx

-4sin(x-y) + 4sin(x-y)*y' = 3sinx*y' + 3ycosx

y'(4sin(x-y) - 3sinx) = 3ycosx + 4sin(x-y)

3ycosx + 4sin(x-y)

----------------------- = y'

4sin(x-y) - 3sinx

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