Tutorial: The derivative: algebraic viewpoint
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Calculating the derivative at a point algebraically
%%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.
$\displaystyle f'(x)$ \t $\displaystyle {}= \lim_{h \to 0} \frac{f(x + h)  f(x)}{h} \qquad$ \t The function $f'$ assigns to each $x$ the value $f'(x)$ of the derivative of $f$ at $x$.
\\ \t !2! ${}={}$Slope of tangent at the point $(x, f(x))$ on the graph, as illustrated here:
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 $x$:
An application: Calculating exact velocity
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).
Calculating the derivative of a rational function
Now let's repeat the method above with a function other than a quadratic function.
A special case: Derivative of x
Let's look at a very special function: $f(x) = x$, whose graph is shown here.

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x + h)  f(x)}{h} = \lim_{h \to 0} \frac{x + h  x}{h} $.
$\displaystyle = \lim_{h \to 0} \frac{(x + h  x)(x+h + x)}{h(x+h + x)} $
\\ $\displaystyle = \lim_{h \to 0} \frac{x + h^2  x^2}{h(x+h + x)} \quad $ \t $\displaystyle (ab)(a+b) = a^2  b^2$
\\ $\displaystyle = \lim_{h \to 0} \frac{(x + h)^2  x^2}{h(x+h + x)} \quad $ \t $a^2 = a^2$ #[for every real number $a$.][para cada número real $a$.]#
\\ $\displaystyle = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 x^2}{h(x+h + x)}$
\\ $\displaystyle = \lim_{h \to 0} \frac{2xh + h^2}{h(x+h + x)}\quad $ \t Simplify.
\\ $\displaystyle = \lim_{h \to 0} \frac{h(2x + h)}{h(x+h + x)}\quad $ \t Factor out the $h$.
\\ $\displaystyle = \lim_{h \to 0} \frac{2x + h}{x+h + x}\quad $ \t Cancel the $h$.
\\ $\displaystyle = \frac{2x}{x + x}\quad $ \t Take the limit; $h = 0$ is now in the domain so long as $x \ne 0$.*
\\ $\displaystyle = \frac{2x}{2x} = \frac{x}{x}\quad $,
* 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 $x$!
%%If $f(x) = x$, %%then $f'(x) = \dfrac{x}{x}\qquad$. \t This is also the same as $\dfrac{x}{x}$, (multiply top and bottom by $x$ to see why) and you can use whichever formulation you prefer.
%%Note
 When $x$ is positive, $x = x$, so $f'(x) = \dfrac{x}{x} = 1$, consistent with the fact that the righthand 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 lefthand 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.
Now try the exercises in Section 3.6 in Applied Calculus or Section 10.6 in Finite Mathematics and Applied Calculus.
Copyright © 2018 Stefan Waner and Steven R. Costenoble