View source for Derivative as a limit

''Derivative'' is the [[slope]] of the [[tangent line]].

But what ''is'' the tangent line? (Silly answer: Line the slope of which is equal to the derivative).

What do we need it for? Recall the [[Preview of calculus: part 1|examples]]:
* Velocity
* Headlights
* Sling
i.e. it is used to study motion.

[[Image:TangentLine.png]]

To find the tangent line, we only need its ''slope''. We define its slope as the ''limit of slopes'' of the [[secant line]]s, lines determined by two points on the graph. For each $x \neq a$, draw a line through $(a,f(a))$ and $(x,f(x))$. We have infinitely many secant lines. Taske one:
$$\text{Slope of this line } = \frac{\text{rise}}{\text{run}} = \frac{f(x) - f(a)}{x-a}$$
Then the slope of the tangent line
$$\lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$

[[Image:SlopeofTangentLine.png|The secant lines keep turning and 2nd is better than first.]]

So, as we move $x$ toward $a$, the secant lines turn and approach the tangent line, provided the limit exists. This number is called ''the derivative of $f$ at $x = a$''. Note: this is not limit of $f$ but of the ''"difference quotient"'', a new function made from $f$.

'''Review exercise.''' Compute and justify:
$$\lim_{x\to \infty} \frac{-x^{2} + 2x - 7}{2x^{2} + 77}$$

'''Example.''' Find the tangent line to $y = x^{2}$ at $(1,1)$.

[[Image:Tangentxsquared.png|Tangent of $y=x^{2}$ at $(1,1)$.]]

Slope = the derivative of $y=x^{2}$ at $x=1$.

Use $f(x) = x^{2}$, $a =1$ in the definition.
$$\begin{aligned}
f^{\prime} &= \lim_{x \to a} \frac{f(x) - f(a)}{x - a} \\
& = \lim_{x \to 1} \frac{x^{2} - 1^{2}}{x - 1} \\
& = \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} \\
& = \lim_{x \to 1} ( x + 1 ) \\
& = 1 + 1 \\
&= 2 \quad \text{ Slope }
\end{aligned}$$

The [[point-slope form]] of the line:
$$ y - 1 = 2(x - 1)$$
Consider:
$$f^{\prime}(a) = \lim_{x \to a} \frac{\overbrace{f(x) - f(a)}^{?}}{\underbrace{x - a}_{0}}$$
What are the limits of the numerator and the denominator?
$$\lim_{x \to a} \underbrace{\left(f(x) - f(a)\right)}_{\text{If continuous, substitute}} = f(a) - f(a) = 0$$
This is an [[indeterminate expression]]. So, possibly the derivative exists.

What if $f$ is ''not'' continuous at $x = a$? Then
$$\lim_{x \to a} \frac{\overbrace{f(x) - f(a)}^{\text{This limit does not exist.}}}{\underbrace{x - a}_{0}}$$
Then $f^{\prime}(a)$ does not exist.

Therefore...

'''Theorem.''' If $f$ is not continuous at $x = a$, then the derivative $f^{\prime}(a)$ does not exist.

We say that $f$ isn't ''[[differentiable]]''.

'''Example.''' Consider
$$f(x) = x^{2} \to f^{\prime}(1) = \lim_{x \to a} \frac{x^{2} - 1^{2}}{x-1}$$
To resolve $\frac{0}{0}$, we factor numerator. It's easy.

Factoring, in general, is hard...

'''Alternative definition of derivative'''

Rename $h = x - 1$. Then

[[Image:AltDerivative.png]]

$$f^{\prime}(a) = \lim_{x \to a} \frac{f(x) - f(a)}{\underbrace{x - a}_{h = x - a}}$$
Since $h = x - a$, then $x = a + h$

Rewrite the definition:
$$ f^{\prime}(a) = \lim_{x \to a} \frac{f(a + h) - f(a)}{h} $$

This makes life easier...

'''Example.''' Given
$$f(x) = x^{2}.$$

$$\begin{aligned}
f^{\prime}(1) &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \\
& = \lim_{h \to 0} \frac{(1 + h)^{2} - 1^{2}}{h} \\
& = \lim_{h \to 0} \frac{1^{2} + 2h + h - 1}{h} \\
& = \lim_{h \to 0} \frac{2h + h^{2}}{h} \qquad \frac{0}{0} \\
& = \lim_{h \to 0} (2 + h^{2}) = 2 + 0^{2} = 2
\end{aligned}$$

[[Image:SecantLineAnimation.png|Animation of secant and tangent lines (http://www.ies.co.jp/math/java/calc/limsec/limsec.html)]]

[http://www.ies.co.jp/math/java/calc/limsec/limsec.html animation site]

'''Problem.''' One can estimate the derivative from the graph. Let's find the derivative at the red point.

[[Image:RedDerivative.png]]

$$\text{Derivative } = \text{ Slope of the tangent line}$$
# Find it. It's the green line.
# Find its slope. $\frac{\text{Rise}}{\text{Run}} = \frac{14}{3} \approx 5$.
To better estimate the slope, draw a triangle as large as possible. 

Here we get algebra from the graph. It's typically vice versa.

Find $f^{\prime}(1)$ for $f(x) = \frac{1}{x}$.

[[Image:Algebra.png]]

$$\begin{aligned}
f^{\prime} &= \lim_{h \to 0} \frac{f(1 + h) - f(1)}{h} \qquad \to \frac{0}{0} \\
&= \lim_{h \to 0} \frac{\frac{1}{1+h} - \frac{1}{1}}{h} \qquad \to \frac{0}{0} \\
&= \lim_{h \to 0} \frac{\frac{1}{1+h} - \frac{1+h}{1+h}}{h} \\
&= \lim_{h \to 0} \frac{1 - (1-h)}{1+h} \cdot \frac{1}{h} \\
&= \lim_{h \to 0} \frac{-h}{1 + h}\cdot \frac{1}{h} \\
&= \lim_{h \to 0} -\frac{1}{1 + h} \\
&= -\frac{1}{1+0} = -1
\end{aligned}$$
This is a rational function, continuous on its domain. To evaluate the limit, plug in.

'''Example.''' Evaluate:
$$\lim_{x \to 5} \frac{2^{x} - 32}{x - 5}. $$
It's just a limit. But we recognize that this is the derivative of some function. Compare to the definition
$$ f^{\prime}(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x - a} $$
and match.

So
$$\begin{gathered}
a = 5 \\
f(x) = 2^{x} \\
f(5) = 2^{5} = 32
\end{gathered}$$

So our limit, $f^{\prime}(5)$ for $f(x) = 2^{x}$. Compute:
$$f^{\prime}(x) = (2^{x})^{\prime} = 2^{x} \ln 2, $$
so
$$f^{\prime}(5) = 2^{5} \ln 2 = 32 \ln 2.$$

[[category:calculus]]