Question

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

Answer 1

Let us consider the graph of `y = 1/x` when `x > 0`. When `x` increases, we know that `y` decreases.

We are given `(dx)/(dt) = "8 min"^(-1)` and are looking for `(dy)/(dt)`, and we already have `y = 1/x` as our equation. Notice how:

`x = x(t)` because the x value varies over time
`y = y(x(t))` because y varies as x varies, and x varies over time

So, we can write `(dy)/(dx) = color(green)((dy(x(t)))/(dx(t)))` and `(dx)/(dt) = color(green)((dx(t))/(dt))`. Therefore, using the chain rule with implicit differentiation to find `(dy)/(dt)`:

`color(blue)((dy)/(dt)) = (dy(x(t)))/(dt)`

`= color(green)((dy(x(t)))/(dx(t))*(dx(t))/(dt))`

`= color(blue)((dy)/(dx)*(dx)/(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)`

`= -1/(2)^2*("8 min"^(-1))`

`= color(blue)(-"2 min"^(-1))`

Answer 2

`y = x^(-1) qquad qquad qquad dy = - x ^(-2) \ dx`

`:. (dy)/(dt) = - 1/x ^( 2) \ (dx)/(dt)`

For `x =2`:

`= - 1/4 *8 = - 2 " min"^(-1)`