Problem. Let $t(x)$ be the equation of the tangent line to the curve $1/x$ at the point $(x, 1/x)$. Let $u(x)$ be the function that maps $x$ to the intersection that $t(x)$ makes with the $y$-axis. Compute the derivative of $u(x)$ at the point $x=1$.
An interpretation of the problem. The problem is asking me to find out how fast the intersection between the $y$-axis and the tangent $t(x)$ is going down as it passes by $x = 1$ as we move $x$ to the right. I say "going down" because I'm using "fast", which sort of implies that $x$ is time and time moves to the right.
A solution. The derivative of $1/x$ is $- 1/x^{2}$. It looks like I need to use a general point $x = a$, otherwise I can't get an algebraic formula for $u(x)$. The slope of the tangent at $(a, 1/a)$ will be $-1/a^2$. Now the tangent line is \begin{align*} t(x) - \frac{1}{a} &= -\frac{1}{a^2}(x - a)\\ t(x) &= -\frac{x}{a^2} + \frac{1}{a} + \frac{1}{a} = \frac{-x + 2a}{a^2}. \end{align*}
It turns out the function $u(x)$ being the function that measures how high $t(x)$ intercepts the vertical axis, we'll end up having $u(a)$ and not $u(x)$ because we'll fix $x$ and let $a$ vary, since the interception happens at $x = 0$. Therefore,
$$u(a) = t(0) = \frac{2a}{a^2} = \frac{2}{a}.$$
I want the derivative of $u(a)$ at $a = 1$,
\begin{align*} u'(a) = \frac{-2}{a^2}. \end{align*}
At $a=1$, we get $u'(1) = -2$. The slope of $u(a)$ looks reasonable because it is always negative and because it goes to zero as $a$ grows without a bound.
Question. How do I verify this is correct?