Answer
(a)$\frac{dy}{dx}=\frac{4x-y-1}{x}$
(b)$\frac{dy}{dx}= 2-\frac{2}{x^2}$
(c)$ \frac{dy}{dx}= 2-\frac{2}{x^2} $
Work Step by Step
$x +xy-2x^2=2$
(a)Find dy/dx by differentiating implicitly.
$ \frac{d}{dx}( x +xy-2x^2)= \frac{d}{dx}(2) $
$ \frac{dx}{dx} +\frac{d(xy)}{dx} -\frac{d(2x^{2})}{dx} =0 $
$1+y \frac{dx}{dx}+x\frac{dy}{dx}-2 \frac{d(x^2)}{dx}=0 $
$1+y+x\frac{dy}{dx}-2(2x)=0$
$1+y+x\frac{dy}{dx}-4x=0$
$x\frac{dy}{dx}=4x-y-1$
$\frac{dy}{dx}=\frac{4x-y-1}{x}$
$(b)x+xy-2x^2=2$
$x+xy=2x^2+2$
$(1+y)x=2x^2+2$
$1+y= \frac{2x^2+2}{x}$
$y= \frac{2x^2+2}{x}-1$
$y=2x+\frac{2}{x}- 1$
$y=2x+2x^{-1}-1$
$\frac{dy}{dx}= \frac{d}{dx}(2x+2x^{-1}-1)$
$\frac{dy}{dx}= \frac{d(2x)}{dx}+\frac{d(2x^{-1})}{dx}-\frac{d(1)}{dx}$
$\frac{dy}{dx}= 2\frac{d(x)}{dx}+2\frac{d(x^{-1})}{dx}-0$
$\frac{dy}{dx}= 2(1)+2(-1x^{-2})$
$\frac{dy}{dx}= 2-\frac{2}{x^2}$
(c)From part (a)
$\frac{dy}{dx}=\frac{4x-y-1}{x}$
$\frac{dy}{dx}=\frac{4x-(y+1)}{x}$
Or
$\frac{dy}{dx}=\frac{4x^2-x(y+1)}{x^2}$ ......... eq(1)
Also
$y+1=\frac{2x^2+2}{x}$
Or
$x(y+1)=2x^2+2$ ....................... eq(2)
From equation (1) and equation (2)
$ \frac{dy}{dx}= \frac{4x^2-2x^2-2}{x^2} $
$ \frac{dy}{dx}= \frac{2x^2-2}{x^2} $
$ \frac{dy}{dx}= 2-\frac{2}{x^2} $
RESULT OF (a) and (b) are the same.
