# What is the implicit derivative of 3=(1-y)/x^2+xy ?

Mar 27, 2017

$y ' = \frac{2 - y \left(2 + {x}^{3}\right)}{- x + {x}^{4}}$

#### Explanation:

First, you should take the derivative of on $3 = \frac{1 - y}{x} ^ 2 + x y$
$\frac{d}{\mathrm{dx}} 3 = \frac{d}{\mathrm{dx}} \frac{1 - y}{x} ^ 2 + \frac{d}{\mathrm{dx}} x y$

Use the Quotient Rule on $\frac{1 - y}{x} ^ 2$
Quotient Rule: $f \frac{x}{g} \left(x\right) = \frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$
$\frac{f ' \left(x\right) g \left(x\right) - g ' \left(x\right) f \left(x\right)}{g \left(x\right)} ^ 2$ = $\frac{\left(- y '\right) \cdot {x}^{2} - 2 x \cdot \left(1 - y\right)}{{x}^{2}} ^ 2$
It is $- y '$ because $\frac{d}{\mathrm{dx}} \left(1 - y\right)$ Derivative of any constant is 0, and the derivative of $- y$, using $\frac{d}{\mathrm{dx}}$ is $\frac{\mathrm{dy}}{\mathrm{dx}}$, which is $- y '$.
Use Product Rule for $\frac{d}{\mathrm{dx}} x \cdot y$
Product Rule: $f \left(x\right) \cdot g \left(x\right) = f ' \left(x\right) \cdot g \left(x\right) + g ' \left(x\right) \cdot f \left(x\right)$
$f ' \left(x\right) \cdot g \left(x\right) + g ' \left(x\right) \cdot f \left(x\right)$ = $\frac{d}{\mathrm{dx}} x \cdot y + \frac{d}{\mathrm{dx}} y \cdot x$
$\frac{d}{\mathrm{dx}} x \cdot y + \frac{d}{\mathrm{dx}} y \cdot x$ = $1 \cdot y + y ' \cdot x = y + y ' x$

Now plug in:
$\frac{d}{\mathrm{dx}} 3 = \frac{d}{\mathrm{dx}} \frac{1 - y}{x} ^ 2 + \frac{d}{\mathrm{dx}} x y$
$0 = \frac{\left(- y '\right) \cdot {x}^{2} - 2 x \cdot \left(1 - y\right)}{{x}^{2}} ^ 2 + y + y ' x$
$0 = \frac{- {x}^{2} y ' - 2 x + 2 x y}{{x}^{4}} + y + y ' x$
$0 = \frac{- y '}{x} ^ 2 - \frac{2}{x} ^ 3 + \frac{2 y}{x} ^ 3 + y + y ' x$
$0 = \frac{- y '}{x} ^ 2 - \frac{2 \left(1 + y\right)}{x} ^ 3 + y + y ' x$

Let's get the denominator to ${x}^{3}$
$0 = \frac{- x y '}{x} ^ 3 - \frac{2 + 2 y}{x} ^ 3 + \frac{{x}^{3} y}{x} ^ 3 + \frac{y ' {x}^{4}}{x} ^ 3$
$0 = \frac{- x y '}{x} ^ 3 + \frac{y ' {x}^{4}}{x} ^ 3 - \frac{2 + 2 y}{x} ^ 3 + \frac{{x}^{3} y}{x} ^ 3$
$0 = \frac{- x y ' + y ' {x}^{4} - 2 + 2 y + {x}^{3} y}{x} ^ 3$
$0 = \frac{y ' \left(- x + {x}^{4}\right) - 2 + y \left(2 + {x}^{3}\right)}{x} ^ 3$
$0 = y ' \left(- x + {x}^{4}\right) - 2 + y \left(2 + {x}^{3}\right)$
$2 - y \left(2 + {x}^{3}\right) = y ' \left(- x + {x}^{4}\right)$
$\frac{2 - y \left(2 + {x}^{3}\right)}{- x + {x}^{4}} = y '$