find y' by implicit differentiation.

Question

i don't understand, its so complicate.

1. 4cos(x-y)=3ysinx
2. 4sin(x-y) =3ysinx

Answer 1

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.

Answer 2

ok i'll try another, thanks

Answer 3

oh
d/dx cos u = - sin u du/dx
You forgot the - sign

Answer 4

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