Implicitly differentiate $xy=1$ $xy = 1  \implies  y = 1/x$, which means that $\mathrm dy/\mathrm dx$ should equal $-x^{-2}$ through the power rule.
How would you get this through implicit differentiation using the equation $xy = 1$?
 A: we get $y+xy'=0$ thus we have $y'=-\frac{y}{x}$ with $x\ne 0$
A: Take the derivative of both sides with respect to x, The Derivative of the right is a derivative of a constant which equals 0 ($\frac{d}{dx}C=0$). The left is a derivative of a product which is $\frac{d}{dx}\left(f\left(x\right)g\left(x\right)\right)=\frac{d}{dx}\left(f\left(x\right)\right)\cdot g\left(x\right)+\frac{d}{dx}\left(g\left(x\right)\right)\cdot f\left(x\right)$. 
So,
$\frac{d}{dx}\left(x\cdot y\right)=\frac{d}{dx}\left(1\right)$
$y+x\cdot \frac{dy}{dx}=0$
Then subtract solve for $\frac{dy}{dx}$
$\frac{dy}{dx}=-\frac{y}{x}$
Finally substitute for y. I think you can take it from their.
A: $$xy=1$$
$$\frac{d}{dx}(xy)=\frac{d}{dx}(1)$$
$$\frac{d}{dx}(x)\cdot y+x\cdot\frac{d}{dx}(y)=\frac{d}{dx}(1)$$
$$1\cdot y+x\cdot\frac{dy}{dx}=0$$
$$x\cdot\frac{dy}{dx}=-y$$
$$\frac{dy}{dx}=-\frac yx$$
$$\frac{dy}{dx}=-\frac{xy}{x^2}=-\frac 1{x^2}$$
A: Differentiate both sides of the equation with respect to $x$:
$$\begin{align}
&\frac{\mathrm d}{\mathrm dx}(xy) = \frac{\mathrm d}{\mathrm dx}1\\[1.3ex]
\implies & y + x\frac{\mathrm dy}{\mathrm dx} = 0\qquad\text{derivative of a product}\\[1.3ex]
\implies & \frac{\mathrm dy}{\mathrm dx} = -\frac yx\\[1.3ex]
\implies & \frac{\mathrm dy}{\mathrm dx} = -\frac1{x^2}\qquad\text{using the fact that } y = \frac1x
\end{align}$$
A: $$
y + xy' = 0
$$
and user similar transformation as in your question
A: From
$xy = 1$,
we get
$x\,dy+y\,dx = 0$
so
$\frac{dy}{dx}
=\frac{-y}{x}
=\frac{-1/x}{x}
=\frac{-1}{x^2}
$.
