Implicitly differentiate $xy=1$

Question

$xy = 1 \implies y = 1/x$, 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 $xy = 1$?

Answer 1

we get $y+xy'=0$ thus we have $y'=-\frac{y}{x}$ with $x\ne 0$

Answer 2

$$ y + xy' = 0 $$ and user similar transformation as in your question

Answer 3

$$xy=1$$ $$\frac{d}{dx}(xy)=\frac{d}{dx}(1)$$ $$\frac{d}{dx}(x)\cdot y+x\cdot\frac{d}{dx}(y)=\frac{d}{dx}(1)$$ $$1\cdot y+x\cdot\frac{dy}{dx}=0$$ $$x\cdot\frac{dy}{dx}=-y$$ $$\frac{dy}{dx}=-\frac yx$$ $$\frac{dy}{dx}=-\frac{xy}{x^2}=-\frac 1{x^2}$$

Answer 4

Differentiate both sides of the equation with respect to $x$: $$\begin{align} &\frac{\mathrm d}{\mathrm dx}(xy) = \frac{\mathrm d}{\mathrm dx}1\\[1.3ex] \implies & y + x\frac{\mathrm dy}{\mathrm dx} = 0\qquad\text{derivative of a product}\\[1.3ex] \implies & \frac{\mathrm dy}{\mathrm dx} = -\frac yx\\[1.3ex] \implies & \frac{\mathrm dy}{\mathrm dx} = -\frac1{x^2}\qquad\text{using the fact that } y = \frac1x \end{align}$$

Answer 5

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.

Answer 6

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} $.