Welcome. So far, we've built a solid foundation with limits. We used limits to handle tricky situations like holes and asymptotes, and to formalize the concept of continuity. Today, we use the limit to solve two major problems that motivated the invention of calculus: the tangent line problem and the instantaneous velocity problem. The solution to both will lead us to the single most important concept in this course: the derivative. 📈
Let's start with a geometric question: how do we find the slope of a line that just skims a curve at a single point? We can't use the old slope formula, $m = (y_2-y_1)/(x_2-x_1)$, because we only have one point. The ingenious idea is to find the slope of a secant line through two nearby points, and then take the limit as those points get infinitely close together.
The slope of the tangent line to the curve $y=f(x)$ at the point $(a, f(a))$ is given by the limit:
$$ m = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $$(Provided this limit exists)
Let's look at each piece of that formula to understand where it comes from:
Why is this limit a big deal? For thousands of years, mathematicians could only describe an average rate of change between two distinct points. The idea of an "instantaneous" rate of change at a single point seemed like a paradox (how can something change at an instant with no time passing?). This limit is the ingenious tool that resolves that paradox. It gives us a mathematically rigorous way to "zoom in" on a single point and determine its properties, allowing us to precisely describe the nature of change itself.
You might wonder why it took until the 17th century for Newton and Leibniz to formalize this, especially when ancient mathematicians like Euclid were using limit-like ideas (the "method of exhaustion") to find areas. The key difference is that the Greek method was used to calculate static, accumulative quantities like area. The revolutionary insight of calculus was to create a system for analyzing motion and rates of change.
This required tools the Greeks didn't have: a fully developed system of algebra, the coordinate geometry of Descartes to link algebra to shapes, and the formal concept of a function. Newton and Leibniz synthesized these ideas into a powerful, general procedure for analyzing how functions change at an instant.
Let $f(x) = x^2 + 2x$.
(a) Find the slope of the tangent line at $x=1$.
(b) Find the equation of that tangent line.
(a) Here, $a=1$ and $f(a) = f(1) = 1^2 + 2(1) = 3$. We use the limit definition for the slope:
$$ \begin{align*} m &= \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} \\ &= \lim_{h \to 0} \frac{[(1+h)^2 + 2(1+h)] - 3}{h} \\ &= \lim_{h \to 0} \frac{(1+2h+h^2) + (2+2h) - 3}{h} \\ &= \lim_{h \to 0} \frac{4h+h^2}{h} \\ &= \lim_{h \to 0} \frac{h(4+h)}{h} = \lim_{h \to 0} (4+h) = 4 \end{align*} $$The slope of the tangent line at $x=1$ is 4.
(b) We have the slope $m=4$ and a point on the line $(1, 3)$. Using the point-slope form $y - y_1 = m(x - x_1)$:
$$ y - 3 = 4(x-1) \implies y = 4x - 4 + 3 \implies y = 4x - 1 $$
Now, let's consider a seemingly different problem from physics. How do we find the velocity of an object at a single instant?
If the position of an object at time $t$ is given by $s(t)$, then the instantaneous velocity at time $t=a$ is given by the limit:
$$ v(a) = \lim_{h \to 0} \frac{s(a+h) - s(a)}{h} $$Notice that the formula for the slope of a tangent line and the formula for instantaneous velocity are mathematically identical! One describes a geometric slope, the other a physical velocity. This powerful, unifying concept is what we will soon call the derivative.
A ball is thrown into the air, and its height (in feet) after $t$ seconds is given by $s(t) = 40t - 16t^2$. What is its instantaneous velocity at $t=2$ seconds?
Here, $a=2$ and $s(2) = 40(2) - 16(2^2) = 80 - 64 = 16$.
$$ \begin{align*} v(2) &= \lim_{h \to 0} \frac{s(2+h) - s(2)}{h} \\ &= \lim_{h \to 0} \frac{[40(2+h) - 16(2+h)^2] - 16}{h} \\ &= \lim_{h \to 0} \frac{(80+40h) - 16(4+4h+h^2) - 16}{h} \\ &= \lim_{h \to 0} \frac{80+40h - 64 - 64h - 16h^2 - 16}{h} \\ &= \lim_{h \to 0} \frac{-24h - 16h^2}{h} \\ &= \lim_{h \to 0} \frac{h(-24 - 16h)}{h} = \lim_{h \to 0} (-24 - 16h) = -24 \end{align*} $$The instantaneous velocity at $t=2$ is -24 ft/s. The negative sign indicates that the ball is moving downwards at that moment.
Problem: Find the equation of the tangent line to the curve $y = 3x - x^2$ at the point $(1, 2)$.
We have now seen that the same limit calculation gives us the slope of a tangent line and the instantaneous rate of change. This fundamental concept is given a special name: the derivative.
The derivative of a function $f$ at a number $a$, denoted by $f'(a)$ (read "f prime of a"), is:
$$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} $$This single number represents both the slope of the tangent line to $y=f(x)$ at $x=a$ and the instantaneous rate of change of $f(x)$ with respect to $x$ at $x=a$.
Find the derivative of the function $f(x) = \sqrt{x}$ at the point $a=4$.
We want to find $f'(4)$.
$$ f'(4) = \lim_{h \to 0} \frac{f(4+h) - f(4)}{h} = \lim_{h \to 0} \frac{\sqrt{4+h} - \sqrt{4}}{h} = \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h} $$This gives $\frac{0}{0}$, so we must use the conjugate method.
$$ \begin{align*} f'(4) &= \lim_{h \to 0} \frac{\sqrt{4+h} - 2}{h} \cdot \frac{\sqrt{4+h} + 2}{\sqrt{4+h} + 2} \\ &= \lim_{h \to 0} \frac{(4+h) - 4}{h(\sqrt{4+h} + 2)} \\ &= \lim_{h \to 0} \frac{h}{h(\sqrt{4+h} + 2)} \\ &= \lim_{h \to 0} \frac{1}{\sqrt{4+h} + 2} = \frac{1}{\sqrt{4+0} + 2} = \frac{1}{2+2} = \frac{1}{4} \end{align*} $$So, $f'(4) = 1/4$. This means the slope of the tangent line to $y=\sqrt{x}$ at $x=4$ is $1/4$.
Problem: Find $f'(3)$ for the function $f(x) = 2x^2 - 5$.
The "prime" notation, $f'(x)$, was developed by the mathematician Joseph-Louis Lagrange. However, another notation, developed by one of the co-inventors of calculus, Gottfried Wilhelm Leibniz, is also extremely common and useful.
Leibniz thought about the slope of the secant line as a ratio of differences. He used $\Delta x$ (delta-x) to represent a small change in $x$ (our $h$), and $\Delta y$ to represent the corresponding change in $y$, so $\Delta y = f(x+\Delta x) - f(x)$. The slope of the secant line is then simply $\frac{\Delta y}{\Delta x}$.
Leibniz reasoned that the derivative is the limit of this ratio as the change in $x$ becomes infinitesimally small. He wrote this as:
$$ \frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} $$This notation is very powerful because it looks like a fraction (though it isn't one!) and is extremely useful in later applications like the chain rule and integration.
The development of calculus in the 17th century was marked by a famous and bitter priority dispute between Sir Isaac Newton in England and Leibniz in Germany. Today, historians agree that both men discovered calculus independently, with Newton likely discovering it first but Leibniz publishing first. We use notation from both traditions today: Newton's dot notation is common in physics, Lagrange's prime notation is clean and simple, and Leibniz's differential notation is incredibly descriptive and powerful.
Find the derivative of $f(x) = x^3 - 4x$ at $a=2$.
Find the derivative of $f(x) = \frac{x}{x+1}$ at $a=1$.
Today, we defined the derivative at a point, $f'(a)$. We've seen that this single number has two powerful interpretations: it is the slope of the tangent line to the graph of $f(x)$ at $x=a$, and it is the instantaneous rate of change of the function at that point. In our next lecture, we will generalize this idea to find the derivative not just at a single point, but as a function, $f'(x)$, that gives us the slope at any point. After that, we will begin developing shortcut rules to save us from having to calculate these long limits every time!