Calculus I Lesson 8: Higher Order Derivatives

  1. Warm Up
    1. x = 0
    2. x = 1 Differentiable?
    3. x = 1 Continuous?
    4. x = 2
    5. Graph
  2. Higher Order Derivatives
    1. Physics Example
    2. Crime Example
    3. COVID-19
  3. Tangent Lines
    1. Graph
    2. Meaning of “Tangent Line”
  4. Reminders

Warm Up

\[f(x) = \begin{cases}x^2 - 1 &\mbox{if } x < 1 \\ x - 1 &\mbox{if } x \geq 1 \end{cases}\]

Is $f(x)$ differentiable, continuous, both, or neither at the following points:

x = 0

$x^2 - 1$ is differentiable at $x = 0$. Therefore it is also continuous.

Answer: Both

x = 1 Differentiable?

Need to check left and right hand limits of

\[\lim_{x \rightarrow 1} \frac{f(x) - f(1)}{x - 1}\]

From the left: $\dfrac{(x^2 - 1) - 0}{x - 1} \rightarrow x + 1 \rightarrow 2$.

From the right: $\dfrac{x -1 - 0}{x - 1} \rightarrow 1$. Not differentiable

x = 1 Continuous?

Now we check if $f(x)$ is continuous at $x = 1$:

So $f$ is continuous at $x = 1$.

x = 2

$x - 1$ is differentiable, its derivative is $1$ near $x = 2$. Differentiable functions are continuous, so both.

Graph

Higher Order Derivatives

Given a function $f$:

Physics Example

Crime Example

Violent Crime Rates Over Time

This graph shows the US Violent Crime Rate over time.

COVID-19

Take a look at the following graphs.

Graph 1: total COVID cases in NY (excluding NYC) over time:

Graph of cumulative number of COVID cases over time

Graph 2: daily COVID cases in NY (excluding NYC) over time:

What does the derivative of each of these graphs represent? What stands out to you about these derivatives?

Tangent Lines

Recall that we defined the derivative to mean: “the slope of the tangent line”. How do we actually figure out what that tangent line is?

Problem: Find the equation of the line tangent to $y = x^3 - 2x$ at $x = 1$.

Recall: point-slope form: $y - y_0 = m(x - x_0)$

So how do we find the equation? First find the point. Since we’re looking at $x = 1$, we plug in to find $y = 1^3 - 2(1) = -1$, so the point is $(1, -1)$.

Now let’s find the slope by taking the derivative. We want the slope at $x = 1$, so we look for $f^\prime(1)$. First find the derivative: $y^\prime = 3x^2 - 2$. At $x = 1$: $3(1)^2 - 2 = 1$.

So our equation is: $y + 1 = 1(x - 1)$, or $y + 1 = x - 1$. Or $y = x- 2$

Graph

Meaning of “Tangent Line”

What does the “tangent line” even mean? The idea is that it’s the best “linear approximation” of $f(x)$ at $x = a$. In other words: zoom in close to $x = a$ on the graph. The closer we get, the more accurate the tangent line appears.

Reminders