Question

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

Answer 1

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

Explanation:

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.!