Posted by josephine on Wednesday, February 29, 2012 at 10:25pm.
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.
ok i'll try another, thanks
oh
d/dx cos u = - sin u du/dx
You forgot the - sign
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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