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

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Answer 1 · 2016-08-18T11:42:29

#dy/dx=-(y-cosy)^2/(2(cosy+ysiny)#.

We rewrite the given eqn. as # -2y=x(cosy-y), or, x=(2y)/(y-cosy)#.

Diff.ing w.r.t. #y# both sides of this eqn., we have,

#dx/dy=d/dy{(2y)/(y-cosy)}=2d/dy{y/(y-cosy)}#.

Using Quotient Rule for Diffn. on the R.H.S., we get,

#dx/dy=2[{(y-cosy)*d/dy(y)-y*d/dy(y-cosy)}/(y-cosy)^2]#

#=2[{(y-cosy)(1)-(y)(1+siny)}/(y-cosy)^2]#

#=2{(y-cosy-y-ysiny)/(y-cosy)^2}#

#=-2{(cosy+ysiny)/(y-cosy)^2}#.

Therefore, #dy/dx=1/(dx/dy)=-(y-cosy)^2/(2(cosy+ysiny)#.

Enjoy Maths.!