Differentiation and Applications

TL;DR

Differentiation helps us find the instantaneous rate of change of a function, which is essentially its slope at any given point. It's a fundamental tool for analyzing how quantities change and finding maximum or minimum values in various real-world scenarios. We'll explore techniques for calculating derivatives and apply them to optimization problems.

1. The Mental Model

Think of differentiation as finding the steepness of a hill at any exact spot, not just between two far-apart points. This steepness tells you how fast something is changing right then.

2. The Core Material

Differentiation is all about finding the derivative of a function, which represents its instantaneous rate of change.

The Derivative as a Limit

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The formal definition of a derivative for a function $f(x)$ at a point $x$ is:

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

This limit essentially calculates the slope of the tangent line to the curve at $x$.

Common Differentiation Rules

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You won't always use the limit definition. Here are some essential rules:

• Constant Rule: If $f(x) = c$ (where $c$ is a constant), then $f'(x) = 0$.
• Power Rule: If $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
• Constant Multiple Rule: If $f(x) = c \cdot g(x)$, then $f'(x) = c \cdot g'(x)$.
• Sum/Difference Rule: If $f(x) = g(x) \pm h(x)$, then $f'(x) = g'(x) \pm h'(x)$.
• Product Rule: If $f(x) = g(x) \cdot h(x)$, then $f'(x) = g'(x)h(x) + g(x)h'(x)$.
• Quotient Rule: If $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$.
• Chain Rule: If $f(x) = g(h(x))$, then $f'(x) = g'(h(x)) \cdot h'(x)$.

Second Derivatives and Concavity

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The second derivative, $f''(x)$, is the derivative of the first derivative. It tells you about the concavity of a function:
* If $f''(x) > 0$, the function is concave up (like a cup).
* If $f''(x) < 0$, the function is concave down (like a frown).
* Points where concavity changes are called inflection points.

Applications: Optimization Problems

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One of the most powerful applications of differentiation is optimization, where we find the maximum or minimum values of a function.

Here's a general approach:

graph TD
    A["Understand the problem & identify quantities"] --> B["Formulate a primary equation (what to optimize)"]
    B --> C{"Is there more than one variable?"}
    C -- Yes --> D["Formulate a secondary equation (constraint)"]
    D --> E["Use constraint to express primary equation in one variable"]
    C -- No --> E
    E --> F["Differentiate the primary equation (find f'(x))"]
    F --> G["Set f'(x) = 0 and solve for critical points"]
    G --> H["Use First or Second Derivative Test to classify critical points"]
    H --> I["Check endpoints (if interval is closed)"]
    I --> J["State the optimal solution (max/min value and location)"]

First Derivative Test:
* If $f'(x)$ changes from positive to negative at a critical point, it's a local maximum.
* If $f'(x)$ changes from negative to positive at a critical point, it's a local minimum.

Second Derivative Test:
* If $f''(c) > 0$ at a critical point $c$, it's a local minimum.
* If $f''(c) < 0$ at a critical point $c$, it's a local maximum.
* If $f''(c) = 0$, the test is inconclusive.

3. Worked Example

Let's find the dimensions of a rectangular garden with the largest possible area if you have 100 meters of fencing.

1. Understand and Identify: We want to maximize the area of a rectangle. We have a constraint on the perimeter.

   • Let length be $L$ and width be $W$.
   • Area $A = L \cdot W$
   • Perimeter $P = 2L + 2W = 100$

2. Primary Equation: $A = L \cdot W$ (what we want to maximize)

3. Secondary Equation (Constraint): $2L + 2W = 100$

4. Express in one variable: From the constraint, $2L = 100 - 2W \implies L = 50 - W$.
   Substitute this into the primary equation:
   $A(W) = (50 - W)W = 50W - W^2$

5. Differentiate: Find $A'(W)$. Using the power rule:
   $A'(W) = 50 - 2W$

6. Set to zero and solve: Find critical points by setting $A'(W) = 0$:
   $50 - 2W = 0 \implies 2W = 50 \implies W = 25$ meters.

7. Classify critical point (Second Derivative Test):
   Find $A''(W)$: $A''(W) = -2$.
   Since $A''(25) = -2 < 0$, this critical point corresponds to a local maximum.

8. Check endpoints: The width $W$ must be between 0 and 50 (since $L = 50-W$, if $W=0$, $L=50$, if $W=50$, $L=0$).
   $A(0) = 0$
   $A(50) = 0$
   The maximum is clearly not at the endpoints.

9. Optimal Solution:
   If $W = 25$ meters, then $L = 50 - 25 = 25$ meters.
   The maximum area is $A = 25 \times 25 = 625$ square meters.
   The garden should be a square with sides of 25 meters.

4. Key Takeaways

• The derivative $f'(x)$ gives you the instantaneous rate of change or the slope of the tangent line at any point $x$.
• Master the common differentiation rules (power, product, quotient, chain) to efficiently compute derivatives.
• The second derivative $f''(x)$ tells you about the concavity of the function (up or down).
• Optimization problems involve finding maximum or minimum values by setting the first derivative to zero.
• Always check critical points and endpoints of the domain when solving optimization problems.
• A positive derivative means the function is increasing; a negative derivative means it's decreasing.
• Local extrema (max/min) occur where the first derivative is zero or undefined.

Common Mistakes to Avoid:
* Forgetting to apply the Chain Rule when differentiating composite functions.
* Incorrectly applying the Product or Quotient Rule – pay close attention to the formula.
* Not considering the domain or endpoints of the function in optimization problems.
* Confusing a zero first derivative with a maximum/minimum without further testing (it could be an inflection point).

5. Now Try It

Find the maximum volume of an open-top box that can be made from a 10 cm by 10 cm square piece of cardboard by cutting equal squares from the corners and folding up the sides. Determine the side length of the squares you should cut out and the maximum volume. What success looks like is having a numerical value for the side length cut out and the corresponding maximum volume of the box.