Tutorial: The derivative: algebraic viewpoint

Adaptive game version

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(This topic is also in Section 3.6 in Applied Calculus or Section 10.6 in Finite Mathematics and Applied Calculus)

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%%Q So far, all we have been doing is approximating the derivative of a function using numerical and graphical approaches. Is there a way of computing it exactly?
%%A When the function is specified algebraically, we can calculate the exact value of the derivative using an algebraic approach, which we show here.

The algebraic approach is quite straightforward: Instead of subtracting numbers to estimate the average rate of change over smaller and smaller intervals, we subtract algebraic expressions using the definition of the derivative at a point $x = a$ in terms of the difference quotient

$\displaystyle f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \qquad$ The derivative is the limit of the difference quotient.

Let's do the calculation by going through an example:

The above calculation gives $f'(x)$ at #[a particular value of $x$][un valor particular de $x$]# #[for the function given. ][para la dada función. ]# However, there is nothing special about this value of $x$. We saw in the %%prevsectutc how to think of the derivative $f'$ as a function: the derivative function:
What we have calculated above is the value of the derivative function $f'$ at a particular $x$. But we can also use the same technique to give us a formula for $f'(x)$ by calculating it at a general $x$:

We saw in the %%prevsectut that, if $f$ is a function of time $t$ (such as, for instance, the height of a launch vehicle) then its derivative $f'$ gives the velocity of $f$ (for instance the velocity of the launch vehicle).

Now let's repeat the method above with a function other than a quadratic function.

Let's look at a very special function: $f(x) = |x|$, whose graph is shown here. Its derivative is

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \lim_{h \to 0} \frac{|x + h| - |x|}{h} $.

Now we use a trick: Multiply top and bottom by the "conjugate": $|x + h| + |x|$, of the numerator: * If $x = 0$, the preceeding expression is $\displaystyle \lim_{h \to 0} \frac{2(0) + h}{|0+h| + |0|} = \lim_{h \to 0}\frac{h}{|h|}$, which does not exist.

and we have found a formula for the derivative of $|x|$! %%Note

• When $x$ is positive, $|x| = x$, so $f'(x) = \dfrac{x}{x} = 1$, consistent with the fact that the right-hand side of the graph above has slope 1.
• When $x$ is negative, $|x| = -x$, and so $f'(x) = \dfrac{x}{-x} = -1$, consistent with the fact that the left-hand side of the graph above has slope $-1$.
• When $x = 0$, $\dfrac{x}{|x|}$ is undefined, and as we saw above, the limit defining the derivative does not exist, meaning that $f'(0)$ does not exist.