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

Question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-10T07:08:12

#y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}#

Given equation:

#-y=\cos y/x-x^2/y#

#y=x^2/y-\cos y/x#

differentiating the above equation w.r.t. #x# on both sides as follows

#\frac{d}{dx}y=\frac{d}{dx}(x^2/y)-\frac{d}{dx}(\cos y/x)#

#y'=\frac{y\frac{d}{dx}x^2-x^2\frac{d}{dx}y}{y^2}-\frac{x\frac{d}{dx}\cos y-\cos y\frac{d}{dx}x}{x^2}#

#y'=\frac{2xy-x^2y'}{y^2}-\frac{-x\siny\ y'-\cos y}{x^2}#

#x^2y^2y'=2x^3y-x^4y'+xy^2\sin y\ y'+y^2\cos y#

#(x^4+x^2y^2-xy^2\sin y)y'=2x^3y+y^2\cos y#

#y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}#