In the previous chapters, we defined the derivative as the limit of a difference quotient. While this definition is the bedrock of differential calculus, calculating derivatives this way can be tedious. Today, our goal is to develop a set of powerful "shortcut" rules that allow us to find derivatives of many common functions quickly and efficiently, bypassing the lengthy limit process. 🏎️
Let's start with the most basic function imaginable: a constant function, $f(x) = c$. Since the graph of this function is a horizontal line, its slope is always zero. This gives us our first, simplest rule.
For any constant $c$, $$\frac{d}{dx}(c) = 0$$
For the function $f(x) = 3$, which is a horizontal line, the derivative is $f'(x) = 0$ for all $x$ in $(-\infty, \infty)$.
Now consider the identity function, $f(x) = x$. Let's find its derivative using the limit definition:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{(x+h) - x}{h} = \lim_{h \to 0} \frac{h}{h} = \lim_{h \to 0} 1 = 1$$
So the derivative is 1. We can write this result in a specific form: if $f(x)=x^1$, then $f'(x) = 1 \cdot x^{1-1}$. Let's see if this holds for $x^2$.
Let's find the derivative of $f(x)=x^2$ using the limit definition.
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{x^2 + 2xh + h^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = \lim_{h \to 0} (2x+h) = 2x$$
So, the derivative of $x^2$ is $2x$. Writing this in the same form as before, we have: if $f(x)=x^2$, then $f'(x) = 2x^{2-1}$.
Looking at these results, a clear pattern seems to be emerging. This suggests a general rule: If $f(x) = x^n$, then the derivative is $f'(x) = nx^{n-1}$. Let's prove that this pattern holds true for any positive integer $n$.
To find the derivative of $f(x) = x^n$ for a positive integer $n$, we use the limit definition:
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^n - x^n}{h}$$
Using the Binomial Theorem to expand $(x+h)^n$:
$$(x+h)^n = x^n + nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n$$
Substituting this back and simplifying:
$$f'(x) = \lim_{h \to 0} \frac{nx^{n-1}h + \frac{n(n-1)}{2}x^{n-2}h^2 + \dots + h^n}{h} = \lim_{h \to 0} (nx^{n-1} + \frac{n(n-1)}{2}x^{n-2}h + \dots + h^{n-1})$$
Every term except the first has an $h$ in it, so as $h \to 0$, all other terms vanish, leaving:
$$f'(x) = nx^{n-1}$$
This powerful result is known as the Power Rule. While we only proved it for positive integers, Isaac Newton and others showed that it remarkably works for any real number $n$.
If $n$ is any real number, then $$\frac{d}{dx}(x^n) = nx^{n-1}$$
Find the derivatives of the following functions:
Find the derivative of $f(x) = \frac{1}{x^2}$.
Now that we can differentiate a power of $x$, we will develop some techniques to differentiate sums and scalar products of these power functions.
If $c$ is a constant and $f$ is a differentiable function, then $$\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]$$ Proof: $\lim_{h \to 0} \frac{c \cdot f(x+h) - c \cdot f(x)}{h} = c \cdot \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = c \cdot f'(x)$.
If $f$ and $g$ are both differentiable, then so are their sum and difference:
$$\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$$
With these rules, we can now differentiate any polynomial, and even more generally, any sum of power functions.
Find the derivative of $P(x) = 4x^3 - 5x^2 + x - 9$.
We apply our rules to each term one-by-one:
$$P'(x) = \frac{d}{dx}(4x^3) - \frac{d}{dx}(5x^2) + \frac{d}{dx}(x^1) - \frac{d}{dx}(9)$$
$$ = 4(3x^2) - 5(2x^1) + (1x^0) - 0$$
$$ = 12x^2 - 10x + 1$$
Let's turn our attention to exponential functions of the form $f(x) = a^x$, where the base $a$ is a positive constant ($a>0$). Let's see what the limit definition tells us.
$$f'(x) = \lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = \lim_{h \to 0} \frac{a^x (a^h - 1)}{h}$$
As we evaluate the limit with respect to $h$, the term $a^x$ does not change. So, we can treat it as a constant and factor it out of the limit:
$$f'(x) = a^x \cdot \left( \lim_{h \to 0} \frac{a^h - 1}{h} \right)$$
Let's pause and look closely at this. The derivative of $a^x$ is the function $a^x$ itself, multiplied by whatever the value of that limit is. But what is that limit? Let's recall the definition of the derivative at a specific point, $x=0$:
$$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}$$
For our function, $f(x) = a^x$, we have $f(0) = a^0 = 1$. Substituting this in gives:
$$f'(0) = \lim_{h \to 0} \frac{a^{0+h} - a^0}{h} = \lim_{h \to 0} \frac{a^h - 1}{h}$$
That mysterious limit is nothing more than the slope of the function at $x=0$. This is a truly cool result!
The derivative of an exponential function is proportional to the function itself, and the constant of proportionality is the slope of the function at $x=0$.
$$\frac{d}{dx}(a^x) = f'(0) \cdot a^x$$
You might ask, "what is $f'(0)$?" The answer is that it depends on the base 'a'. You might then ask a more interesting question: "What if $f'(0) = 1$?" This question is the key to everything. Phrased differently: for the function $f(x) = a^x$, for which value of 'a' is the derivative at zero exactly equal to 1? The answer defines the most important number in calculus. (We will come back and find a formula for $f'(0)$ for any base 'a' later in the course!)
The number $e \approx 2.71828$ is the unique base for which the slope of the tangent line at $x=0$ is exactly 1.
$$\lim_{h \to 0} \frac{e^h - 1}{h} = 1$$
Explore this concept in the Desmos graph below. Adjust the slider for the base 'a' and watch the slope of the tangent line at $x=0$. Notice that the slope equals 1 only when 'a' is the number e. At this point, the derivative graph (in green) perfectly overlaps the original function (in blue).
Because of this special property, the derivative rule for $f(x) = e^x$ is the simplest non-zero derivative rule of all.
$$\frac{d}{dx}(e^x) = e^x$$
The function $f(x)=e^x$ is its own derivative!
Find the derivative of $f(x) = e^x - 5x^2$.
We use the difference rule and apply our new rule for $e^x$ and the power rule for $x^2$.
$$f'(x) = \frac{d}{dx}(e^x) - \frac{d}{dx}(5x^2) = e^x - 5(2x) = e^x - 10x$$
Find the derivative of $f(x) = 4e^x - 9x + 15$.
A hobby rocket is launched upward from the ground. Its height in feet after $t$ seconds is given by the position function $s(t) = -16t^2 + 128t$.
a) Find the function $v(t)$ that describes the rocket's instantaneous velocity at any time $t$.
b) When does the rocket reach its maximum height? (Hint: What is the velocity at the exact moment it stops going up?)
Find the derivative of $h(x) = 2e^x + 7x^5 - \pi^3$.
Today, we have built a powerful toolkit for finding derivatives without direct use of the limit definition. We established rules for constants, powers, constant multiples, and sums/differences, which together allow us to differentiate any sum of power functions. We then investigated the unique properties of exponential functions, discovering the special number $e$ and the elegant rule that $\frac{d}{dx}(e^x) = e^x$. These rules are foundational, and we will use them in almost every subsequent topic in calculus. Next time, we will learn how to find derivatives of products and quotients of functions. ✅