How do you differentiate #y=xcosy^2-xy#?

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Answer 1 · 2016-10-21T15:46:56

#dy/dx=(cosy^2-y)/(1+2xysiny^2+x)#

#y=xcosy^2-xy#

We can differentiate each component and then put it together

#y'=dy/dx#

The next one is a chain differentiation and differentiation of a product

#(xcosy^2)'=1*cosy^2 +x(-siny^2)2ydy/dx#
#= cosy^2-(2xysiny^2)dy/dx#

The next one is #(xy)'=1*y+xdy/dx#

Putting it together

#dy/dx=cosy^2-(2xysiny^2)dy/dx-y-xdy/dx#

#dy/dx(1+2xysiny^2+x)=cosy^2-y#

#dy/dx=(cosy^2-y)/(1+2xysiny^2+x)#