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Section 2.1 Intro to Derivatives

Derivatives are first used as a way to describe rates of change. Consider how your position changes as you drive. We can think about \(\dfrac{\text{change in position}}{\text{change in time}}\text{,}\) which looks like a slope:

Figure 2.1.1.

\begin{equation*} \Delta p = 47 - 10 = 37 \text{ miles} \end{equation*}
\begin{equation*} \Delta t = 30 - 0 = 30 \text { minutes} \end{equation*}
\begin{equation*} \dfrac{\Delta p}{\Delta t} = \dfrac{37}{30} \cdot \dfrac{1 \text{hour}}{60 \text{minutes}} = 74 \text{mph (average speed)} \end{equation*}

All ways of getting there give the same result, the average speed.

We can contrast average speed with instantaneous speed. Think of the average as over an interval and instantaneous as at a point. To get an instantaneous speed, we take a limit as the length of the interval between points shrinks to zero.

Consider \(f(t) = t^2\) (pretend for \(t \geq 0\text{,}\) f(t) describes the position of some object):

Figure 2.1.2.

  • what is the average speed of the object between \(t = 0\) and \(t = 1\text{?}\)

    \begin{equation*} \text{Avg.} = \dfrac{f(1) - f(0)}{1 - 0} = \dfrac{1 - 0}{1 - 0} = 1 \text{ m/s} \end{equation*}

  • what is the average speed of the object between \(t = 1\) and \(t = 2\text{?}\)

    \begin{equation*} \text{Avg.} = \dfrac{f(2) - f(1)}{2 - 1} = \dfrac{4 - 1}{2 - 1} = 3 \text{ m/s} \end{equation*}

We call these secant lines.

Definition 2.1.3. Secant Lines.

Lines that cut through the graph of a function at 2 points on the graph.

The slope refers to the average rate of change of function values over the inverval. The slope of the secant, \(M_{\text{sec}} = \dfrac{\Delta f(x)}{\Delta x}\)

Subsection 2.1.1 Secant and Tangent Lines

Figure 2.1.4. Secant Line

slope is \(M_{\text{sec}} = \dfrac{f(b)- f(a)}{b - a}\)

In contrast to a secant lines that cuts a function, a tangent lines touches the graph of the function at a point of interest.

Figure 2.1.5. Tangent Line at \(x = a\)

\(M_{\text{tan}} = \displaystyle\lim_{\Delta x \to 0} \dfrac{f(a - \Delta x) - f(a)}{\Delta x}\) gives the slope of the tangent line at \(x = a\) and is the instantaneous rate of change at \(x = a\text{.}\)

Subsection 2.1.2 Define Derivative

The function \(f'(x)\) is defined by:

\begin{equation*} f'(x) = \lim_{h \to 0} \dfrac{f(x + h) - f(x)}{h} \end{equation*}

is called the derivative of \(f(x)\) with respect to \(x\text{.}\)

This function describes the instantaneous rate of change of \(f(x)\) at each point on the graph (provided the limit exists).

Example 2.1.1.

What is the slope (tangent line) of the function \(f(x) = x^2 + 4\) at \(x = 2\) ?

\begin{align*} M_{\text{tan}} \amp = \lim_{\Delta x \to 0} \dfrac{f(2 + \Delta x) - f(2)}{\Delta x}\\ \amp = \lim_{\Delta x \to 0} \dfrac{(2 + \Delta x)^2 + 4 - (2^2 + 4)}{\Delta x}\\ \amp = \lim_{\Delta x \to 0} \dfrac{2^2 + 4 \Delta x + (\Delta x)^2 + 4 - 2^2 - 4}{\Delta x}\\ \amp = \lim_{\Delta x \to 0} \dfrac{4 \Delta x + (\Delta x)^2 }{\Delta x}\\ \amp = \lim_{\Delta x \to 0} \dfrac{\Delta x (4 + \Delta x)}{\Delta x}\\ \amp = \lim_{\Delta x \to 0} 4 + \Delta x\\ \amp = 4 + 0 = 4 \end{align*}

To create the equation to find a tangent line of a function we use the point slope form:

\begin{equation*} y = y_0 + m(x - x_0) \end{equation*}

where \((x_0,y_0)\) is a point on a line and \(m\) is a slope.

So the equation for the tangent line is:

\begin{align*} f_{\text{tan}}(x) \amp = f(x_0) + M_{\text{tan}}(x - x_0)\\ \amp = f(2) + M_{\text{tam}}(x - 2)\\ \amp = 8 + 4(x - 2) \end{align*}

\(f(2) = 2^2 + 4 = 8\)

\(M_{\text{tan}}\) at \(x = 2\) is 4 (as shown above)

In general, we are interested in \(f'(x)\) defined by \(f'(x) = \displaystyle\lim_{ h \to 0} \dfrac{f(x + h)- f(x)}{h}\)

Recall, \(f(x) = x^2 + 4\)

\begin{align*} f'(x) \amp = \lim_{h \to 0} \dfrac{f(x + h) - f(2)}{h}\\ \amp = \lim_{h \to 0} \dfrac{(x + h)^2 + 4 - (x^2 + 4)}{h}\\ \amp = \lim_{h \to 0} \dfrac{x^2 + 2xh + h^2 + 4 - x^2 + 4}{h}\\ \amp = \lim_{h \to 0} \dfrac{2xh + h^2}{h}\\ \amp = \lim_{h \to 0} \dfrac{h(2x + h)}{h}\\ \amp = \lim_{h \to 0} 2x + h\\ \amp = 2x + 0 = 2x \end{align*}

So, for \(f(x) = x^2 + 4\text{,}\) \(f'(x) = 2x\text{.}\) The slope of \(f(x)\) at various points is given by values of \(f'(x)\) at those points. \(f'(x) = 2x \) is the derivative of \(f(x) = x^2 + 4\text{.}\)

Using the derivative of \(f(x) = x^2 + 4\text{,}\) we can verify the slopes of the tangent lines to \(f(x)\) found earlier.

  • at \(x =2\text{,}\) the slope of the tangent line is \(M_{\text{tan}} = f'(2) = 2 \cdot 2 = 4\)

  • at \(x = 0\text{,}\) the slope of the tangent line is \(M_{\text{tan}} = f'(0) = 2 \cdot0 = 0\)

Big Idea.

We often are interested in approximating a function by it's tangent line; this makes solving certain hard/impossibe problems easier/possible.

The process of finding derivatives is called differentiation. The following are interchangeable:

  • \(f'(x) = \dfrac{df}{dx} = \dfrac{d}{dx} \left( f(x) \right)\) for the derivative

  • \(f'(x_0) = \dfrac{df}{dx} |_{x = x_{0}} = \dfrac{d}{dx} \left( f(x) \right) |_{x = x_{0}}\) for the derivative at a given point

Power Rule.

For \(f(x) = x^{n}\text{,}\) \(f'(x) = \dfrac{d}{dx} \left( x^n \right) = nx^{n-1}\)

For square roots, etc., the equation needs to be converted: \(f(x) = \sqrt{x} = x^{\frac{1}{2}}\text{.}\)

We say functions like \(f(x) = x^2 + 4\) are differentialble, meaning we can fully apply and work out the derivative by the limit definition (and the limit in the definition exists).

Some functions are not differentiable at certain points. There are a few simple reasons why a function can be indifferentiable at a point:

  • it has a corner at that point (ex: the aboslute value function); there are infinite ways to draw the langent line at a corner

  • when the tangent line would be verticle; this would be an undefined slope

  • if the function is discontinuous at that point