d/dx ( sin^2 x y cos x y = 1)

Question

Find the derivative of the following functions:

Answer 1

I'm assuming this is sin²(xy)*cos(xy) = 1   So you'll need the product rule and the chain rule.

 

sin²(xy) *[-sin(xy)*(x dy/dx) + y)] + cos(xy)[2sin(xy)cos(xy) *(x dy/dx + y) = 0 simplifying you'd get

 

-xsin³(xy) dy/dx - ysin³⁽xy) + 2xsin(xy)cos²(xy) dy/dx + 2ysin(xy)cos²(xy) = 0 bring all non dy/dx terms on other side

 

-xsin³(xy) dy/dx + 2xsin(xy)cos²(xy) dy/dx  =  ysin³(xy) - 2ysin(xy)cos²(xy)    factor out dy/dx

 

dy/dx (-xsin³(xy) + 2xsin(xy)cos²(xy)          = ysin³(xy) - 2ysin(xy)cos²(xy)    and lastly, divide to get dy/dx

 

dy/dx = [ ysin³(xy) - 2ysin(xy)cos²(xy) ] / [-xsin³(xy) + 2xsin(xy)cos²(xy) ] factor out -y/x, the rest cancels

 

dy/dx = -y/x

 

Hope this helps.

Answer 2

 Find d/dx of sin²xycosxy = 1

 

First use chain rule and product rule on the left side of the equation.

 

-ysinxysin²xy + 2ycosxycosxy = 0 

 

2ycos²xy  =   ysin³xy

 

2cos²xy   =  sin³xy

 

2 = tan²xysinxy