Posted by josephine on .
find y' by implicit differentiation.
i don't understand, its so complicate.
1. 4cos(xy)=3ysinx
2. 4sin(xy) =3ysinx

Calculus implicit 
David,
4sin(xy)*d(xy)=3y' + 3ycosx
On the left I applied the chain rule, and on the right i used the product rule.
4sin(xy)*(1y')=3y' + 3ycosx
Differentiate x  y and get 1  y'
Then do algebra to solve for y'
4sin(xy)  y'*4sin(xy) = 3y' + 3ycosx
3y' + y'*4sin(xy) = 4sin(x2)  3ycosx
y' = [4sin(x2)3ycosx]/[3+4sin(xy)]
Try the other one yourself. 
Calculus implicit 
josephine,
ok i'll try another, thanks

Calculus implicit 
Damon,
oh
d/dx cos u =  sin u du/dx
You forgot the  sign 
Calculus implicit  oops 
Steve,
also missed on the right side :(
4cos(xy)=3ysinx
4sin(xy)(1y') = 3sinx*y' + 3ycosx
4sin(xy) + 4sin(xy)*y' = 3sinx*y' + 3ycosx
y'(4sin(xy)  3sinx) = 3ycosx + 4sin(xy)
3ycosx + 4sin(xy)
 = y'
4sin(xy)  3sinx