What is the implicit derivative of 1=x/y ?

May 4, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x}$

Since $y = x$, $\frac{\mathrm{dy}}{\mathrm{dx}} = 1$

Explanation:

We have $f \left(x , y\right) = \frac{x}{y} = 1$

$\frac{x}{y} = x {y}^{-} 1$

We first derivate with respect to $x$ first:
$\frac{d}{\mathrm{dx}} \left[x {y}^{-} 1\right] = \frac{d}{\mathrm{dx}} \left[1\right]$

${y}^{-} 1 + x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = 0$

Using the chain rule, we get:
$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

${y}^{-} 1 + \frac{\mathrm{dy}}{\mathrm{dx}} x \frac{d}{\mathrm{dx}} \left[{y}^{-} 1\right] = 0$

${y}^{-} 1 + \frac{\mathrm{dy}}{\mathrm{dx}} - x {y}^{-} 2 = 0$

$\frac{\mathrm{dy}}{\mathrm{dx}} x {y}^{-} 2 = {y}^{-} 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{-} \frac{1}{x {y}^{-} 2} = {y}^{2} / \left(x y\right) = \frac{y}{x}$

Since, we know $y = x$ we can say that $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{x} = 1$