Answer
a) $\frac{dy}{dx} = \frac{2-y-3x^{2}}{x}$
b) $\frac{dy}{dx} = -\frac{1}{x^{2}}-2x$
c) $\frac{dy}{dx} = -\frac{1}{x^{2}}-2x$
Work Step by Step
a) $x^{3}+xy-2x = 1$
Differentiating implicitly :
$3x^{2}+x\frac{dy}{dx}+y-2 = 0$
$x\frac{dy}{dx} = -3x^{2}-y+2$
$\frac{dy}{dx} = \frac{2-y-3x^{2}}{x}$
b) $y= \frac{1+2x-x^{3}}{x}$
$y = \frac{1}{x} + \frac{2x}{x} - \frac{x^{3}}{x}$
$y = x^{-1}+2-x^{2}$
Differentiating :
$\frac{dy}{dx} = -\frac{1}{x^{2}}-2x$
c) $\frac{dy}{dx} = \frac{2-(\frac{1}{x}+2-x^{2})-3x^{2}}{x}$
Simplifying :
$\frac{dy}{dx} = \frac{2x-(1+2x-x^{3})-3x^{3}}{x^{2}}$
$\frac{dy}{dx} = \frac{-1+x^{3}-3x^{3}}{x^{2}}$
$\frac{dy}{dx} = \frac{-1-2x^{3}}{x^{2}}$
$\frac{dy}{dx} = -\frac{1}{x^{2}}-2x$
