Calculus Essentials

Derivatives & Optimization

The Linear Model

For this demonstration, download the grades.csv dataset.

d <- read.csv('grades.csv')

head(d)
  midterm final overall gradeA
1   79.25 47.00    69.2      0
2   96.25 87.75    94.3      1
3   58.25 37.75    62.0      0
4   54.50 62.00    72.4      0
5   83.00 39.75    72.4      0
6   41.75 49.50    59.5      0

The Linear Model

plot(d$midterm, d$final, 
     xlab = 'Midterm Grade', 
     ylab = 'Final Grade')

The Linear Model

m <- lm(final ~ midterm, data = d) # predict final grade from midterm grade

abline(a = m$coefficients['(Intercept)'], b = m$coefficients['midterm'])

The Linear Model

\[ y_i = \alpha + \beta x_i + \varepsilon_i \]

The Linear Model

Partitioning the outcome into two parts—the part we can explain, and the part we’re ignoring:

\[ \underbrace{y_i}_\text{outcome} = \underbrace{\alpha + \beta x_i}_\text{explained} + \underbrace{\varepsilon_i}_\text{unexplained} \]

The Linear Model

Partitioning the outcome into two parts—the part we can explain, and the part we’re ignoring:

\[ \underbrace{y_i}_\text{outcome} = \overbrace{\alpha}^\text{intercept parameter} + \beta x_i + \varepsilon_i \]

The Linear Model

Partitioning the outcome into two parts—the part we can explain, and the part we’re ignoring:

\[ \underbrace{y_i}_\text{outcome} = \overbrace{\alpha}^\text{intercept parameter} + \underbrace{\beta}_\text{slope parameter} x_i + \varepsilon_i \]

The Linear Model

Partitioning the outcome into two parts—the part we can explain, and the part we’re ignoring:

\[ \underbrace{y_i}_\text{outcome} = \overbrace{\alpha}^\text{intercept parameter} + \underbrace{\beta}_\text{slope parameter} \overbrace{x_i}^\text{explanatory variable} + \varepsilon_i \]

The Linear Model

Partitioning the outcome into two parts—the part we can explain, and the part we’re ignoring:

\[ \underbrace{y_i}_\text{outcome} = \overbrace{\alpha}^\text{intercept parameter} + \underbrace{\beta}_\text{slope parameter} \overbrace{x_i}^\text{explanatory variable} + \underbrace{\varepsilon_i}_\text{prediction error} \]

. . .

But where do the \(\alpha\) and \(\beta\) values come from? How do we estimate the “line of best fit”?

An Optimization Problem

We want to find values for \(\alpha\) and \(\beta\) that minimize the sum of squared error.

. . .

sse <- function(a,b){
  y <- d$final # outcome
  x <- d$midterm # explanatory variable
  
  predicted_y <- a + b*x
  
  error <- y - predicted_y
  
  return( sum(error^2) )
}

An Optimization Problem

plot(d$midterm, d$final,
     xlab = 'Midterm Grade', ylab = 'Final Grade')

abline(a = 10, b = 0.5) # too shallow

sse(a = 10, b = 0.5)
[1] 54632.59

An Optimization Problem

plot(d$midterm, d$final,      
     xlab = 'Midterm Grade', ylab = 'Final Grade')  

abline(a = 0, b = 1.2) # too steep!

sse(a = 0, b = 1.2)
[1] 61043.95

An Optimization Problem

We could keep hunting blindly for values \(\alpha\) and \(\beta\) that minimize the sum of squared errors, or we could take a more systematic approach…

. . .

\[ \text{SSE} = \sum_{i=1}^n(y_i - \alpha - \beta x_i)^2 \]

An Optimization Problem

\(\text{SSE} = \sum_{i=1}^n(y_i - \alpha - \beta x_i)^2\)

. . .

Imagine dropping a ball on this surface. The ball will roll until it reaches a perfectly flat point: the function’s minimum.

Review: Slopes

What is the slope of this function? \(f(x) = 3x + 2\)

. . .

The slope of a linear function (a straight line) is measured by how much \(y\) increases when you increase \(x\) by \(1\). In this case, \(3\).

Review: Slopes

Find the slope of each function:

  • \(y = 2x + 4\)

  • \(f(x) = \frac{1}{2}x - 2\)

  • life expectancy (years) = 18.09359 + 5.737335 \(\times\) log(GDP per capita)

. . .

Slope of a line \(= \frac{rise}{run} = \frac{\Delta Y}{\Delta X} = \frac{f(x+h) - f(x)}{h}\)

Finding the slope of a line is easy.

Just wanted you to get comfortable with that last expression, because we’ll be see it again in a moment.

https://smartech.gatech.edu/bitstream/handle/1853/56031/effect_of_gdp_per_capita_on_national_life_expectancy.pdf

Nonlinear Functions

. . .

Nonlinear functions are confusing and scary…

Nonlinear functions are confusing and scary. Sometimes the slope is positive. Sometimes it’s negative. Sometimes it’s zero. And unlike with linear functions, just looking at the formula gives you no indication what the slope is at any point.

Newton & Leibniz

Developed/Discovered: - The theory of universal gravitation - Three Laws of Motion - The Nature of Light - And, as a side project so he’d have mathematical notation for those other projects, he created calculus

(NB Fall 2020: Newton did some of his best work while stuck at home during an epidemic. So, you know, get to it.)

Also invented calculus, but with better notation. A philosophical optimist who believed we lived in the “best of all possible worlds”, a sentiment parodied by Voltaire, and perhaps belied by the fact that Newton took all the credit for inventing calculus.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

. . .

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The Key Insight

Any curve becomes a straight line if you “zoom in” far enough.

The point is that, in the limit, as you shrink the interval smaller and smaller (infinitesimally small), the function is better and better approximated by a straight line. And we already know the slope of a straight line, so the problem is solved! (That line is called the tangent line FYI.)

The Key Insight

Putting all that into math…

\[ f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \]

Putting all that into math…

\[ f'(x) = \underbrace{\lim_{h \to 0}}_\text{shrink h really small}\frac{\overbrace{f(x+h)-f(x)}^\text{the change in y}}{\underbrace{h}_\text{the change in x}} \]

. . .

This is called the derivative of a function. Using the derivative, you can find the slope at any point.

Derivative Example

Let \(f(x) = 2x + 3\). What is \(f'(x)\)?

. . .

\[ f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h} \]

. . .

\[ = \lim_{h \to 0}\frac{2(x+h)+3-(2x+3)}{h} \]

. . .

\[ = \lim_{h \to 0}\frac{2x+2h+3-(2x+3)}{h} \]

Derivative Example

Let \(f(x) = 2x + 3\). What is \(f'(x)\)?

\[ = \lim_{h \to 0}\frac{2x+2h+3-(2x+3)}{h} \]

. . .

\[ = \lim_{h \to 0}\frac{2h}{h} \]

. . .

\[ = 2 \]

Hey look what we just discovered! The slope of a linear function equals the coefficient on \(x\)!

Now A Nonlinear Example

Let \(f(x) = 3x^2 + 2x + 3\). What is \(f'(x)\)?

. . .

\[ = \lim_{h \to 0}\frac{3(x+h)^2 + 2(x+h) + 3 - (3x^2 + 2x + 3)}{h} \]

. . .

\[ = \lim_{h \to 0}\frac{3x^2 + 3h^2 + 6xh + 2x+ 2h + 3 - (3x^2 + 2x + 3)}{h} \]

\[ = \lim_{h \to 0}\frac{3h^2 + 6xh + 2h}{h} \]

Now A Nonlinear Example

Let \(f(x) = 3x^2 + 2x + 3\). What is \(f'(x)\)?

\[ = \lim_{h \to 0}\frac{3h^2 + 6xh + 2h}{h} \]

. . .

\[ = \lim_{h \to 0}3h + 6x + 2 \]

. . .

\[ = 6x + 2 \]

Solution

The function \(f'(x)\), outputs the slope of \(f(x)\) at every point.

Derivative Shortcuts

Good news! You don’t have to go through that process every time. Mathematicians have done it for you, and have discovered a whole bunch of useful shortcuts.

Shortcut 1: The Power Rule

If \(f(x) = ax^k\), then \(f'(x) = kax^{k-1}\)

. . .

Example: If \(f(x) = 5x^4\), then \(f'(x) = 20x^3\).

. . .

Practice Problem: Let \(f(x) = 2x^3\). What is \(f'(x)\)?

. . .

\[f'(x) = 6x^2\]

Shortcut 2: The Sum Rule

The derivative of a sum is equal to the sum of derivatives.

If \(f(x) = g(x) + h(x)\), then \(f'(x) = g'(x) + h'(x)\)

. . .

Example: If \(f(x) = x^3 + x^2\), then \(f'(x) = 3x^2 + 2x\)

. . .

Practice Problem: If \(f(x) = 2x^3 + x^2\), what is \(f'(x)\)?

. . .

\[f'(x) = 6x^2 + 2x\]

Shortcut 3: The Constant Rule

The derivative of a constant is zero.

If \(f(x) = c\), then \(f'(x) = 0\)

. . .

Example: If \(f(x) = 5\), then \(f'(x) = 0\).

. . .

Practice Problem: If \(f(x) = 4x^2 + 3x + 5\), what is \(f'(x)\)?

. . .

\[ f'(x) = 8x + 3 \]

Shortcut 4: The Product Rule

The derivative of a product is a bit trickier…

If \(f(x) = g(x) \cdot h(x)\), then \(f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)\)

. . .

Example: If \(f(x) = (2x)(x + 2)\), then \(f'(x) = 2x + 2(x+2) = 4x + 4\)

. . .

Practice Problem: \(f(x) = (3x^2 + 6x)(x+2)\), what is \(f'(x)\)?

. . .

\[f'(x) = (3x^2 + 6x)(1) + (6x + 6)(x+2)\]

\[f'(x) = 3x^2 + 6x + 6x^2 + 6x + 12x + 12\]

\[f'(x) = 9x^2 + 24x + 12\]

Shortcut 5: The Chain Rule

If \(f(x) = g(h(x))\), then \(f'(x) = g'(h(x)) \cdot h'(x)\)

. . .

“The derivative of the outside times the derivative of the inside.”

. . .

Example: If \(f(x) = (2x^2 - x + 1)^3\), then \(f'(x) = 3(2x^2 - x + 1)^2 (4x - 1)\)

. . .

Practice Problem: \(f(x) = \sqrt{x + 3} = (x+3)^{\frac{1}{2}}\), what is \(f'(x)\)?

. . .

\(f'(x) = \frac{1}{2}(x+3)^{-\frac{1}{2}}(1) = \frac{1}{2\sqrt{x+3}}\)

More Practice

  • Let \(f(x) = 2x^3 + 4x + 79\). What is \(f'(x)\)?
  • Let \(f(x) = 3(x^2 + x + 42)\). What is \(f'(x)\)?
  • Let \(f(x) = (x^2 + 1)(x+3)\). What is \(f'(x)\)?

Now We Can Do Optimization!

Let \(f(x) = 2x^2 + 8x - 32\). At what value of \(x\) is the function minimized?

Key Insight: Function is minimized when the slope “switches” from decreasing to increasing. Exactly at the point where the slope equals zero.

Optimization in Three Steps

. . .

1. Take the derivative of the function.

. . .

2. Set it equal to zero.

. . .

3. Solve for \(x\).

Optimization in Three Steps

1. Take the derivative of the function.

\[ f(x) = 2x^2 + 8x - 32 \]

. . .

\[ f'(x) = 4x + 8 \]

. . .

2. Set it equal to zero

. . .

\[ 4x + 8 = 0 \]

. . .

3. Solve for \(x\).

. . .

\[ x = -2 \]

That second step is called the “First Order Condition”, or FOC.

Optimization in Three Steps

Now You Try It!

Suppose that happiness as a function of jellybeans consumed is \(h(j) = -\frac{1}{3}j^3 + 81j + 2\). How many jellybeans should you eat? (Assume you can only eat a positive number of jellybeans).

Now You Try It!

Suppose that happiness as a function of jellybeans consumed is \(h(j) = -\frac{1}{3}j^3 + 81j + 2\). How many jellybeans should you eat? (Assume you can only eat a positive number of jellybeans).

Wait.

How do you know if it’s a maximum or a minimum?

. . .

\(h(j) = \frac{1}{3}j^3 + 81j + 2\) and \(h'(j) = 81 - j^2\)

Wait.

It’s a maximum when the slope is decreasing, and a minimum when then slope is increasing. How do you figure out if the slope is increasing or decreasing?

. . .

That’s right. You find the slope of the slope (the derivative of the derivative, aka the second derivative).

The Second Derivative Test

\(h(j) = \frac{1}{3}j^3 + 81j + 2\) and \(h'(j) = 81 - j^2\)

What is \(h''(j)\)? Is it positive or negative when you eat \(9\) jellybeans?

The Second Derivative Test

\(h(j) = \frac{1}{3}j^3 + 81j + 2\) and \(h'(j) = 81 - j^2\)

What is \(h''(j)\)? Is it positive or negative when you eat \(9\) jellybeans? \[ h''(j) = -2j \]

The Second Derivative Test

Partial Derivatives

. . .

What if you have a multivariable function?

\[ f(x,y) = 2x^2y + xy - 4x + y -6 \]

. . .

Same procedure! To get the derivative of a function with respect to \(x\) or \(y\), treat the other variable as a constant.

. . .

\[ \frac{\partial f}{\partial x} = 4yx + y - 4 \]

. . .

\[ \frac{\partial f}{\partial y} = 2x^2 + x + 1 \]

Now You Try!

Suppose happiness as a function of jellybeans and Dr. Peppers consumed is

\[h(j,d) = 8j -\frac{1}{2}j^2 + 2d - 3d^2 + jd + 100\]

How many jellybeans should you eat? How many Dr. Peppers should you drink?

Intuitively, the \(jd\) term is an interaction effect. The effect of jellybeans on happiness increases if you also drink more Dr. Peppers.

Now You Try!

\[ h(j,d) = 8j -\frac{1}{2}j^2 + 2d - 3d^2 + jd + 100 \]

. . .

\[ \frac{\partial h}{\partial j} = 8 - j + d = 0 \]

\[ \frac{\partial h}{\partial d} = 2 - 6d + j = 0 \]

. . .

\[ j = 8 + d \]

\[ j = 6d - 2 \]

\[ d^* = 2 \]

\[ j^* = 10 \]

\[h(j,d) = 8j -\frac{1}{2}j^2 + 2d - 3d^2 + jd + 100\]

The Payoff…

We finally have the tools we need to find the values of \(\alpha\) and \(\beta\) that minimize this function:

\(\text{SSE} = \sum_{i=1}^n(y_i - \alpha - \beta x_i)^2\)