Posted by josephine on Wednesday, February 29, 2012 at 10:25pm.
find y' by implicit differentiation.
i don't understand, its so complicate.
1. 4cos(xy)=3ysinx
2. 4sin(xy) =3ysinx

Calculus implicit  David, Wednesday, February 29, 2012 at 10:43pm
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, 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(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
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