Derivatives are the engine of calculus. They measure how fast things change — and once you master the core rules, you can differentiate almost any function you’ll ever encounter in Calculus 1. This lesson walks through every major rule and technique, with clear explanations and worked examples for each one.
📚 Table of Contents
- What is a Derivative?
- The Limit Definition
- Power Rule
- Basic Derivative Rules
- Product Rule
- Quotient Rule
- Chain Rule
- Derivatives of Trig Functions
- Derivatives of Exponential & Log Functions
- Implicit Differentiation
- Higher Order Derivatives
- Common Mistakes to Avoid
- Practice Problems
1. What is a Derivative?
A derivative measures the instantaneous rate of change of a function. Geometrically, it gives you the slope of the tangent line to the curve at any given point.
Think of it this way: if f(x) describes your position, then f'(x) describes your velocity — how fast your position is changing at any instant.
Notation
All of these mean the same thing — the derivative of f with respect to x:
$$f'(x) \qquad \frac{dy}{dx} \qquad \frac{d}{dx}[f(x)] \qquad \dot{y}$$
2. The Limit Definition of the Derivative
Before using the shortcut rules, it’s important to understand where they come from. The formal definition is:
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) – f(x)}{h}$$
This is the slope formula between two points on the curve, as the distance between them shrinks to zero.
✏️ Example 1 — Using the Definition
Problem: Find f'(x) for \( f(x) = x^2 \) using the limit definition.
Step 1: Write out the formula:
$$f'(x) = \lim_{h \to 0} \frac{(x+h)^2 – x^2}{h}$$
Step 2: Expand:
$$= \lim_{h \to 0} \frac{x^2 + 2xh + h^2 – x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h}$$
Step 3: Factor and cancel:
$$= \lim_{h \to 0} \frac{h(2x + h)}{h} = \lim_{h \to 0} (2x + h) = 2x$$
✅ Answer: \( f'(x) = 2x \)
3. Power Rule
The Power Rule is the most used rule in all of Calculus 1. Learn it cold.
$$\frac{d}{dx}[x^n] = nx^{n-1}$$
Bring the exponent down as a coefficient, then subtract 1 from the exponent. That’s it.
✏️ Example 2
Problem: Differentiate \( f(x) = x^5 \)
$$f'(x) = 5x^4$$
✏️ Example 3
Problem: Differentiate \( f(x) = x^{-3} \)
$$f'(x) = -3x^{-4}$$
✏️ Example 4
Problem: Differentiate \( f(x) = \sqrt{x} = x^{1/2} \)
$$f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$$
✏️ Example 5
Problem: Differentiate \( f(x) = \frac{1}{x^4} = x^{-4} \)
$$f'(x) = -4x^{-5} = \frac{-4}{x^5}$$
4. Basic Derivative Rules
These rules let you handle sums, differences, and constants without breaking a sweat.
| Rule | Formula | Example |
|---|---|---|
| Constant Rule | $$\frac{d}{dx}[c] = 0$$ | $$\frac{d}{dx}[7] = 0$$ |
| Constant Multiple Rule | $$\frac{d}{dx}[cf(x)] = cf'(x)$$ | $$\frac{d}{dx}[5x^3] = 15x^2$$ |
| Sum Rule | $$\frac{d}{dx}[f+g] = f’+g’$$ | $$\frac{d}{dx}[x^2+x^3] = 2x+3x^2$$ |
| Difference Rule | $$\frac{d}{dx}[f-g] = f’-g’$$ | $$\frac{d}{dx}[x^4-x] = 4x^3-1$$ |
✏️ Example 6 — Combining Rules
Problem: Differentiate \( f(x) = 4x^3 – 7x^2 + 2x – 9 \)
$$f'(x) = 12x^2 – 14x + 2$$
5. Product Rule
When two functions are multiplied together, you can’t just differentiate each one separately. Use the Product Rule:
$$\frac{d}{dx}[f(x) \cdot g(x)] = f'(x)g(x) + f(x)g'(x)$$
A helpful way to remember it: “derivative of first times second, plus first times derivative of second.”
✏️ Example 7
Problem: Differentiate \( f(x) = x^2 \cdot \sin(x) \)
Let \( f = x^2 \), \( g = \sin(x) \), so \( f’ = 2x \), \( g’ = \cos(x) \)
$$\frac{d}{dx} = 2x\sin(x) + x^2\cos(x)$$
✏️ Example 8
Problem: Differentiate \( f(x) = (3x + 1)(x^2 – 4) \)
Let \( f = 3x+1 \), \( g = x^2-4 \), so \( f’ = 3 \), \( g’ = 2x \)
$$\frac{d}{dx} = 3(x^2-4) + (3x+1)(2x) = 3x^2 – 12 + 6x^2 + 2x = 9x^2 + 2x – 12$$
6. Quotient Rule
When one function is divided by another, use the Quotient Rule:
$$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) – f(x)g'(x)}{[g(x)]^2}$$
Memory trick: “Low d-High minus High d-Low, over Low squared.”
✏️ Example 9
Problem: Differentiate \( f(x) = \frac{x^2 + 1}{x – 3} \)
Let \( f = x^2+1 \), \( g = x-3 \), so \( f’ = 2x \), \( g’ = 1 \)
$$f'(x) = \frac{2x(x-3) – (x^2+1)(1)}{(x-3)^2} = \frac{2x^2 – 6x – x^2 – 1}{(x-3)^2} = \frac{x^2 – 6x – 1}{(x-3)^2}$$
✏️ Example 10
Problem: Differentiate \( f(x) = \frac{\sin(x)}{x} \)
$$f'(x) = \frac{\cos(x) \cdot x – \sin(x) \cdot 1}{x^2} = \frac{x\cos(x) – \sin(x)}{x^2}$$
7. Chain Rule
The Chain Rule handles composite functions — a function inside another function. It’s the most important rule in Calculus 1.
$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$$
In plain English: differentiate the outside function (leaving the inside alone), then multiply by the derivative of the inside.
✏️ Example 11
Problem: Differentiate \( f(x) = (3x + 2)^5 \)
Outside: \( (\cdot)^5 \) → Inside: \( 3x+2 \)
$$f'(x) = 5(3x+2)^4 \cdot 3 = 15(3x+2)^4$$
✏️ Example 12
Problem: Differentiate \( f(x) = \sin(x^2) \)
Outside: \( \sin(\cdot) \) → Inside: \( x^2 \)
$$f'(x) = \cos(x^2) \cdot 2x = 2x\cos(x^2)$$
✏️ Example 13
Problem: Differentiate \( f(x) = e^{3x^2} \)
Outside: \( e^{(\cdot)} \) → Inside: \( 3x^2 \)
$$f'(x) = e^{3x^2} \cdot 6x = 6xe^{3x^2}$$
✏️ Example 14 — Chain Rule + Product Rule
Problem: Differentiate \( f(x) = x^2 \cdot (x^3 + 1)^4 \)
Use Product Rule with Chain Rule on the second factor:
$$f'(x) = 2x(x^3+1)^4 + x^2 \cdot 4(x^3+1)^3 \cdot 3x^2$$
$$= 2x(x^3+1)^4 + 12x^4(x^3+1)^3$$
$$= 2x(x^3+1)^3\left[(x^3+1) + 6x^3\right]$$
$$= 2x(x^3+1)^3(7x^3+1)$$
8. Derivatives of Trig Functions
These six derivatives must be memorized. They appear constantly on exams.
| Function | Derivative |
|---|---|
| $$\sin(x)$$ | $$\cos(x)$$ |
| $$\cos(x)$$ | $$-\sin(x)$$ |
| $$\tan(x)$$ | $$\sec^2(x)$$ |
| $$\csc(x)$$ | $$-\csc(x)\cot(x)$$ |
| $$\sec(x)$$ | $$\sec(x)\tan(x)$$ |
| $$\cot(x)$$ | $$-\csc^2(x)$$ |
✏️ Example 15
Problem: Differentiate \( f(x) = 3\sin(x) – 2\cos(x) \)
$$f'(x) = 3\cos(x) + 2\sin(x)$$
✏️ Example 16 — Chain Rule with Trig
Problem: Differentiate \( f(x) = \sin(5x^2) \)
$$f'(x) = \cos(5x^2) \cdot 10x = 10x\cos(5x^2)$$
✏️ Example 17
Problem: Differentiate \( f(x) = \tan^3(x) = [\tan(x)]^3 \)
$$f'(x) = 3\tan^2(x) \cdot \sec^2(x)$$
9. Derivatives of Exponential & Log Functions
| Function | Derivative |
|---|---|
| $$e^x$$ | $$e^x$$ |
| $$a^x$$ | $$a^x \ln(a)$$ |
| $$\ln(x)$$ | $$\frac{1}{x}$$ |
| $$\log_a(x)$$ | $$\frac{1}{x\ln(a)}$$ |
✏️ Example 18
Problem: Differentiate \( f(x) = e^{5x} \)
$$f'(x) = e^{5x} \cdot 5 = 5e^{5x}$$
✏️ Example 19
Problem: Differentiate \( f(x) = \ln(x^2 + 1) \)
$$f'(x) = \frac{1}{x^2+1} \cdot 2x = \frac{2x}{x^2+1}$$
✏️ Example 20
Problem: Differentiate \( f(x) = 2^x \)
$$f'(x) = 2^x \ln(2)$$
10. Implicit Differentiation
When y is not isolated on one side of the equation, you can’t differentiate directly. Instead, differentiate both sides with respect to x, treating y as a function of x and applying the Chain Rule whenever you differentiate a term with y.
Every time you differentiate a y term, multiply by \( \frac{dy}{dx} \).
✏️ Example 21
Problem: Find \( \frac{dy}{dx} \) for \( x^2 + y^2 = 25 \)
Step 1: Differentiate both sides:
$$2x + 2y\frac{dy}{dx} = 0$$
Step 2: Solve for \( \frac{dy}{dx} \):
$$2y\frac{dy}{dx} = -2x \Rightarrow \frac{dy}{dx} = \frac{-x}{y}$$
✏️ Example 22
Problem: Find \( \frac{dy}{dx} \) for \( x^3 + y^3 = 6xy \)
Step 1: Differentiate both sides:
$$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$
Step 2: Collect \( \frac{dy}{dx} \) terms:
$$3y^2\frac{dy}{dx} – 6x\frac{dy}{dx} = 6y – 3x^2$$
Step 3: Factor and solve:
$$\frac{dy}{dx}(3y^2 – 6x) = 6y – 3x^2 \Rightarrow \frac{dy}{dx} = \frac{6y – 3x^2}{3y^2 – 6x} = \frac{2y – x^2}{y^2 – 2x}$$
11. Higher Order Derivatives
You can differentiate a function more than once. The second derivative measures how the rate of change is itself changing — in physics, this is acceleration.
| Order | Notation | Meaning |
|---|---|---|
| 1st | \( f'(x) \) or \( \frac{dy}{dx} \) | Rate of change / slope |
| 2nd | \( f”(x) \) or \( \frac{d^2y}{dx^2} \) | Concavity / acceleration |
| 3rd | \( f”'(x) \) or \( \frac{d^3y}{dx^3} \) | Jerk (rate of acceleration) |
✏️ Example 23
Problem: Find \( f”(x) \) for \( f(x) = 4x^4 – 3x^3 + 2x \)
$$f'(x) = 16x^3 – 9x^2 + 2$$
$$f”(x) = 48x^2 – 18x$$
12. Common Mistakes to Avoid ⚠️
- ❌ Forgetting the Chain Rule: \( \frac{d}{dx}[\sin(x^2)] \neq \cos(x^2) \) — you must multiply by \( 2x \)
- ❌ Using Product Rule on a constant: \( \frac{d}{dx}[5x^3] = 15x^2 \), not a product rule situation
- ❌ Quotient Rule sign error: It’s always top derivative times bottom minus top times bottom derivative — order matters!
- ❌ Forgetting \( \frac{dy}{dx} \) in implicit differentiation: Every y term needs it
- ❌ Derivative of \( e^x \) with chain rule: \( \frac{d}{dx}[e^{3x}] = 3e^{3x} \), not just \( e^{3x} \)
- ❌ Power Rule on exponentials: \( \frac{d}{dx}[e^x] \neq xe^{x-1} \) — the Power Rule only applies to \( x^n \)
13. Practice Problems 📝
Try these before checking the answers!
1.
$$\frac{d}{dx}[6x^4 – 3x^2 + 7]$$
👁️ Show Solution
$$24x^3 – 6x$$
2.
$$\frac{d}{dx}[x^3 \cdot e^x]$$
👁️ Show Solution
$$3x^2 e^x + x^3 e^x = x^2 e^x(3 + x)$$
3.
$$\frac{d}{dx}\left[\frac{x^2 + 3}{2x – 1}\right]$$
👁️ Show Solution
$$\frac{2x(2x-1) – (x^2+3)(2)}{(2x-1)^2} = \frac{4x^2 – 2x – 2x^2 – 6}{(2x-1)^2} = \frac{2x^2 – 2x – 6}{(2x-1)^2}$$
4.
$$\frac{d}{dx}[(x^2 + 5)^6]$$
👁️ Show Solution
$$6(x^2+5)^5 \cdot 2x = 12x(x^2+5)^5$$
5.
$$\frac{d}{dx}[\ln(3x^2 + 1)]$$
👁️ Show Solution
$$\frac{6x}{3x^2+1}$$
6. Find \( \frac{dy}{dx} \) for \( x^2 + 3y^2 = 12 \)
👁️ Show Solution
$$2x + 6y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = \frac{-x}{3y}$$
🔗 Continue Your Calculus 1 Journey
More lessons coming soon — check back for the full Calculus 1 series.



