# If (dx)/(dt) = "8 min"^(-1) and y = 1/x, then what is (dy)/(dt) at x = 2?

Oct 15, 2015

Let us consider the graph of $y = \frac{1}{x}$ when $x > 0$. When $x$ increases, we know that $y$ decreases.

We are given $\frac{\mathrm{dx}}{\mathrm{dt}} = {\text{8 min}}^{- 1}$ and are looking for $\frac{\mathrm{dy}}{\mathrm{dt}}$, and we already have $y = \frac{1}{x}$ as our equation. Notice how:

$x = x \left(t\right)$ because the x value varies over time
$y = y \left(x \left(t\right)\right)$ because y varies as x varies, and x varies over time

So, we can write $\frac{\mathrm{dy}}{\mathrm{dx}} = \textcolor{g r e e n}{\frac{\mathrm{dy} \left(x \left(t\right)\right)}{\mathrm{dx} \left(t\right)}}$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = \textcolor{g r e e n}{\frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}}}$. Therefore, using the chain rule with implicit differentiation to find $\frac{\mathrm{dy}}{\mathrm{dt}}$:

$\textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dt}}} = \frac{\mathrm{dy} \left(x \left(t\right)\right)}{\mathrm{dt}}$

$= \textcolor{g r e e n}{\frac{\mathrm{dy} \left(x \left(t\right)\right)}{\mathrm{dx} \left(t\right)} \cdot \frac{\mathrm{dx} \left(t\right)}{\mathrm{dt}}}$

$= \textcolor{b l u e}{\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}}$

So, now let's actually take the derivative with respect to $t$, and then plug in $x = 2$:

color(blue)((dy)/(dt)) = {:[-1/x^2(dx)/(dt)]|:}_(x=2)

$= - \frac{1}{2} ^ 2 \cdot \left({\text{8 min}}^{- 1}\right)$

$= \textcolor{b l u e}{- {\text{2 min}}^{- 1}}$

Jul 31, 2018

$y = {x}^{- 1} q \quad q \quad q \quad \mathrm{dy} = - {x}^{- 2} \setminus \mathrm{dx}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dt}} = - \frac{1}{x} ^ \left(2\right) \setminus \frac{\mathrm{dx}}{\mathrm{dt}}$

For $x = 2$:

$= - \frac{1}{4} \cdot 8 = - 2 {\text{ min}}^{- 1}$