Suppose we have a curve $y=f(x)$. To get the equation of the line tangent to our curve at $(a,f(a))$, we need to

  1. Figure out the slope of the tangent line. This is $$m = f'(a)=\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}.$$

  2. Use the point-slope formula $y-y_0 = m(x-x_0)$ to get the equation of the line: $$ y - f(a) = m (x-a).$$

[Warning: $f'(a)$ is a number, not a function of $x$! If you compute the derivative using a formula, you have to plug in $x=a$.]

Once we have the equation of this tangent line, we can use it to approximate the function near $x=a$. This is useful both for predicting the future, and for infering things about the past. In the following video, we work this out for the example we considered on the previous page, namely $s(t)=4t^2+3$.

As we computed earlier, $s'(2)=16$ and $s(2)=19$. The equation of a line through $(2,19)$ with slope 16 is then \begin{eqnarray*} s-19 &=& 16 (t-2), \hbox{ or} \cr s &=& 19 + 16(t-2), \hbox{ or} \cr s &=& 16t - 13. \end{eqnarray*} You should recognize this as the microscope equation.

All three forms of the equation are equivalent. The third form probably looks simplest, since it involves $t$ rather than $t-2$, but that actually makes it less useful for calculations where $t$ is close to $2$. The second form is better. If we wanted $s(2.01)$ we would take
$19 + 16(2.1-2) = 19 + 16(0.01) = 19 + 0.16 = 19.16$,
and if we wanted $s(1.99)$ we would take
$19 + 16(1.99-2) = 19 + 16(-0.01) = 19 - 0.16= 18.84$.
That's both simpler and less likely to generate errors than computing $16(2.01)-13$ or $16(1.99)-13$.