How do you implicitly differentiate #2= xy-sin^2x-xcos^2y #?

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Answer 1 · 2018-05-13T09:08:53

The answer is #=-((y-sin(2x)-cos^2y)/(x+xsin(2y)))#

Let,

#f(x,y)=xy-sin^2x-xcos^2y-2#

And

#dy/dx=-((delf) /(delx))/((delf) /(dely))#

#(delf) /(delx)=y-2sinxcosx-cos^2y#

#(delf) /(dely)=x+2xcosysiny#

Therefore,

#dy/dx=-(y-2sinxcosx-cos^2y)/(x+2xcosysiny)#

Rearranging,

#dy/dx=-((y-sin(2x)-cos^2y)/(x+xsin(2y)))#