How do you differentiate #sin^2x/y^2-sin^2y/x^2=16#?

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Answer 1 · 2016-07-31T16:44:11

#(dy)/(dx)=(xsin^2x+2x^2sinxcosx-16xy^2)/(16x^2y+ysin^2y+y^2siny cosy)#

#sin^2x/y^2-sin^2y/x^2=16# and multiplying both sides by #x^2y^2#

#hArrx^2sin^2x-y^2sin^2y=16x^2y^2#

Now differentiating both sides, with respect to #x#, note whenever we have a function of #y#, we differentiate w.r.t. #y# and multiply by #(dy)/(dx)#

#2x xxsin^2x+x^2xx2sinx xxcosx-2ysin^2yxx(dy)/(dx)-y^2xx2siny xxcosyxx(dy)/(dx)=16xx2x xxy^2+16x^2xx2yxx(dy)/(dx)# or

#2xsin^2x+4x^2sinxcosx-2ysin^2y(dy)/(dx)-2y^2siny cosy(dy)/(dx)=32xy^2+32x^2y(dy)/(dx)# or

#32x^2y(dy)/(dx)+2ysin^2y(dy)/(dx)+2y^2siny cosy(dy)/(dx)=2xsin^2x+4x^2sinxcosx-32xy^2# or

#(dy)/(dx)=(2xsin^2x+4x^2sinxcosx-32xy^2)/(32x^2y+2ysin^2y+2y^2siny cosy)#

= #(xsin^2x+2x^2sinxcosx-16xy^2)/(16x^2y+ysin^2y+y^2siny cosy)#