Volume 3, Why Slopes and More Math.
Example 1. Let x=2. Then with $x_1=x=2$ and
$x_2 =x_1 +\Delta x =x +\Delta x,$ we have
\begin{eqnarray*} \Delta y & =& f(x_{2}) -f(x_{1}) \\
& =& f(x+\Delta x) -f(x) \\ & =& f(2+\Delta x)- f(2) \\
& =& (2+\Delta x)^2 -2^2 \\ & =& (2^2+2(2)\Delta x+(\Delta
x)^2) -2^2 \\ & =& 2(2)\Delta x+(\Delta x)^2 \end{eqnarray*}
Of course $2(2)=4$, but for the sake of pattern recognition and emphasis,
we keep the arithmetic expression $2(2)$ to the end of the calculation.
Now
\begin{eqnarray*} A& =& \frac{2(2)\Delta x+(\Delta
x)^2)}{\Delta x} \\ & =& 2(2)+(\Delta x) \end{eqnarray*}
This implies
\[ \lim_{\Delta x \to 0}A =\lim_{\Delta x \to 0} 2(2)+(\Delta
x)= 2(2) =4 \]
Example 2. Let x=3. Then
\begin{eqnarray*} \Delta y & =& f(x+\Delta x) -f(x) \\
& =& f(3+\Delta x)- f(3) \\ & =& (3+\Delta x)^2 -3^2 \\
& =& (3^2+2(3)\Delta x+(\Delta x)^2) -3^2 \\
& =& 2(3)\Delta x+(\Delta x)^2 \end{eqnarray*}
Therefore
\begin{eqnarray*} A& =& \frac{2(3)\Delta x+(\Delta x)^2)}{\Delta x} \\ & =& 2(3)+(\Delta x) \end{eqnarray*}
This implies
\[ \lim_{\Delta x \to 0}A =\lim_{\Delta x \to 0}2(3)+(\Delta x)=
2(3) =6 \]
Example 3. Let x=5. Then
\begin{eqnarray*} \Delta y & =& f(x+\Delta x) -f(x) \\
& =& f(5+\Delta x)- f(5) \\ & =& (5+\Delta x)^2 -5^2 \\
& =& (5^2+2(5)\Delta x+(\Delta x)^2) -5^2 \\
& =& 2(5)\Delta x+(\Delta x)^2 \end{eqnarray*}
Therefore we expect
\begin{eqnarray*} A& =& \frac{2(5)\Delta x+(\Delta
x)^2)}{\Delta x} \\ & =& 2(5)+(\Delta x) \end{eqnarray*}
The last equality suggests that
\[ \lim_{\Delta x \to 0}A =\lim_{\Delta x \to 0}2(5)+(\Delta x)
= 2(5) =10 \]
[Play
Video] 2.25 minutes: Derivative as a Limit of a Quotient.
First pass at finding the derivative or slope of f(x) =
x2. Algebraic View.
[Play
Video] 2.25 minutes: Second pass at finding the derivative or
slope of f(x) = x2 at two values of x. Numerical
Examples of Limit Evaluation to suggest a
pattern.
[Play
Video] 3.75 minutes: Third pass at finding the derivative or
slope of f(x) = x2. Back to the algebraic view and a
conclusion.
The Common Algebraic Pattern
The three examples follow the same pattern. We will rewrite the above
calculations with the letter $a$ replacing the numbers 2, 3
and/or 5 above, to emphasize the pattern. In the rewrite below,
note that the role of A below could be played or assumed by each of
the numbers 2, 3 or 5 above, another number or another
letter!
Example n. Let x=a. Then as before
\begin{eqnarray*} \Delta y & =& f(x+\Delta x) -f(x) \\
& =& f(a+\Delta x)- f(a) \\ & =& (a+\Delta x)^2 -a^2 \\
& =& (a^2+2a\Delta x+(\Delta x)^2) -a^2 \\ & =& 2a\Delta
x+(\Delta x)^2 \end{eqnarray*}
Therefore
\begin{eqnarray*} A& =& \frac{2a\Delta x+(\Delta
x)^2)}{\Delta x} \\ & =& 2(a)+(\Delta x) \end{eqnarray*}
This implies
\[ \lim_{\Delta x \to 0}A=\lim_{\Delta x \to 0}2(a)+(\Delta x) =
2a \]
(Note that in the limit calculation, the variable a is held
constant while $\Delta x \to 0$.
Now we can replace a in the above pattern by x. This yields
\[ f'(x)=\lim_{\Delta x \to 0}A = \cdots = \lim_{\Delta x\to0}
2x+(\Delta x) =2x \]
The $\cdots$ indicates reasoning similar or identical to that has
gone before.
Remark. The ratio \[A=\frac{f(x_{1}+\Delta
x)-f(x_{1})}{\Delta x}\] is not defined at $\Delta x =0$ as division by
zero is not defined. But the algebraic manipulations above shows that
$\lim_{\Delta x \to 0} A$ does exist at least for the simple cases
treated.
[Play
Video] 2.25 minutes: Derivative of x3
algebraically via Limits.