Now that you have mastered the rules and formulas of differentiation, it is time for us to apply them! This section focuses on taking all the previous derivative rules and applying them in different contexts. ⚡️

🔍 Prerequisite Information

In order to be able to successfully understand this section, you need to know the difference between an average rate of change and an instantaneous rate of change. 🫡

This section requires you to know how to differentiate using

• Power Rule
• Product Rule
• Quotient Rule
• Chain Rule
• Implicit Differentiation

It also requires you to know how to differentiate:

• Trigonometric Functions
• Inverse Trigonometric Functions
• Inverse Functions
• Logarithmic Functions
• Exponential Functions

Besides all of the calculus prerequisite information, we will also be working on using some formulas you would have learned in previous courses, such as formulas for:

• Area
• Volume
• Perimeter
• Circumference
• Pythagorean Theorem

Source: Giphy

😯 What does a derivative actually mean?

As we know from previous units, a derivative is the instantaneous rate of change of a function, and it is also the slope of the tangent line of a function. For the most part, we have used the xy-plane to compute derivatives. But this is not the only option! By the end of this unit, you should have a good understanding of how to use derivatives to solve problems in many other scenarios! 🤙

📐 One-Dimensional Motion

In AP Calculus, we only learn how to solve problems in 2 dimensions, which means that we only look at motion moving along one axis. The other dimension will be time. ⌚

How might we model this problem? Suppose the position of the particle is given by some function of time $x(t)$. If you just wanted to find the average velocity of the particle over some time interval $t_1$ to $t_2$, we just calculate the change in position (given by the position function) and divide it by the change in time, to get $v_{\text{avg}}=\frac{\Delta x}{\Delta t}$. From this intuition, you might be able to guess that the instantaneous velocity (over an infinitesimal change in time) will be $\frac{dx}{dt}$.

The instantaneous velocity will always be signed. We usually define right as the positive direction and left as the negative direction, so if we get that $\frac{dx}{dt} = -3$ for some value of $t$, we know that the particle is moving at speed 3 to the left. If we want just the speed, then we use the magnitude of the velocity, which for one-dimensional motion is $|v| = 3$.

If we just know the velocity at some instant, we can’t really tell if the particle is speeding up or slowing down. To answer that question, we need to know the acceleration. The acceleration is the rate of change of velocity, $\frac{dv}{dt}$, meaning that if velocity is increasing over time, there is positive acceleration, and if velocity is decreasing over time, there is negative acceleration. What if the acceleration is 0? That means that the particle is moving at a constant rate. ➡️

To summarize:

where $x(t)$ is the position function of the object, $v$ is the velocity, and, $a$ is the acceleration.

🔀 Related Rates

We use related rates when the rate of one thing happening is dependent on the rate of another thing happening. A very common example is considering how fast the volume or area of an object changes if the radius, height, etc. is changing at a certain rate. The best way to learn how to do related rates problems is to just do a lot of them! Every problem will be a little different, and the challenge is in modeling the problem — not necessarily doing the calculations. ✍️

Here is an easy example to get a feel for what you will have to do. Suppose you are blowing a bubble, which is perfectly spherical, and the radius of the bubble increases at a constant rate of 15 mm/s. How fast is the volume increasing? ⁉️

Since we want to find how fast the volume is increasing, we want to find $\frac{dV}{dt}$. Now, $V = 4\pi r^2$, and $\frac{dr}{dt} = 15$, so we have:

Now, we might be asked how much the volume of the bubble is increasing after blowing it for 2 seconds. We will assume that when $t=0$, the bubble has radius 0, so when $t=2$, the bubble will have radius 30 mm. So, the volume is increasing at $3600\pi \text{ mm}^2/\text{s} = 36\pi\text{ cm}^2/\text{s}$.

There are many other examples of related rates problems. Don’t be scared if the function has more than one variable! Try not to overcomplicate things — keep the variable that you need and rewrite the missing variable in terms of the variables you have. For example, if you want to find $\frac{dV}{dt}$ for a cone and you are given $\frac{dr}{dt}$, but not $\frac{dh}{dt}$, then you can find $h$ in terms of $r$ and $V$. See this guide for an example! 👈

📍 Approximating Functions

Sometimes, you may come across scenarios where you want to approximate a complicated function using a much similar linear function. This comes up a lot in statistics and machine learning. The TL;DR here is that if you are given a linear function $y$ that is tangent to some function $f$ at $(x_0, y_0)$, then for some small range of $x$-values around $x_0$, you can approximate $f(x)$ by plugging $x$ into $y$. 👌

You might, however, be asked whether you are underestimating or overestimating the function. If the function is concave up at $x_0$, then the tangent line will be below $f$, so you will get an underestimate. Likewise, if it is concave down at $x_0$, you will overestimate $f$. 🤓

🏩 L’Hopitals Rule

You may remember that when finding limits of functions, there would be some limits that would evaluate to $0/0$ or $\pm \infty/\infty$ unless you were able to cancel terms. L’Hopital’s rule is an easy trick to find these limits. 🧸

L’Hopital’s Rule:

if the following necessary conditions are met: 4️⃣

1. $\displaystyle \lim_{x \to c}f(x) = \lim_{x \to c}g(x) = 0$ or $\pm \infty$
2. $f(x)$ and $g(x)$ are differentiable on an open interval $\mathcal{I}$ except for possibly at point $c \in \mathcal{I}$. This means that both functions are differentiable everywhere around c, but they may or may not be differentiable around c.
3. $\displaystyle \lim_{x \to c}g'(x)\ne0$ , and
4. $\displaystyle \lim_{x \to c}{f'(x)}/{g'(x)}$ exists

Some of these conditions seem obvious, while others may not be. In order to get some intuition about why we might need for these conditions to be met, we will show L’Hospital’s rule in the case where $\displaystyle \lim_{x \to c}f(x) = \lim_{x \to c}g(x) = 0$. If we assume all of the other necessary conditions, then

Now, we can see that if $\displaystyle \lim_{x \to c}f(x) \ne \lim_{x \to c}g(x)$, L’Hopital’s probably won’t work, and if $f(x)$ and $g(x)$ are not differentiable around c, then this method also won’t work. 🙅

A useful corollary to L’Hopital’s rule is that if $f$ is a function that is continuous at $a$ and $f'(x)$ exists for all values in an open interval containing $a$ (except for maybe $x=a$), then if $\displaystyle \lim_{x \to a}f'(x)$ exists, then