AP Calculus BC Unit 2 Notes: Basic Differentiation Rules (Definition to Toolkit)

Derivative Linearity: Constants, Sums, Differences, and Constant Multiples

A derivative measures how fast a function’s output changes as its input changes. If $y = f(x)$, then $f'(x)$ (read “$f$ prime of $x$”) is the instantaneous rate of change of $f$ with respect to $x$. In AP Calculus, you learn the derivative from the limit definition first, but very quickly you switch to differentiation rules that let you compute derivatives efficiently.

The first group of rules are sometimes called linearity rules because they tell you derivatives “distribute” nicely over addition and scalar multiplication. These rules matter because most functions you meet are built by combining simpler functions, and you want a method that lets you differentiate piece-by-piece rather than restarting from the limit definition every time.

Notation you’ll see (all mean the same derivative)

Constant Rule

If a function is a constant, its output never changes as $x$ changes, so its rate of change is zero.

If $f(x) = c$ where $c$ is a constant, then:

$f'(x) = 0$

Why this makes sense: a horizontal line has slope $0$ everywhere.

Example 1

Let $f(x) = 7$. Then:

$f'(x) = 0$

Example 2 (constant inside a larger expression)

If $g(x) = x^2 + 5$, the derivative of the $+5$ part is $0$, so the constant doesn’t affect the rate of change.

Constant Multiple Rule

If you scale a function by a constant, you scale its rates of change by the same constant. Think of stretching a graph vertically: all slopes get stretched by that factor too.

If $f(x)$ is differentiable and $c$ is a constant:

$\frac{d}{dx}[c f(x)] = c f'(x)$

Example 1

Differentiate $h(x) = 4x^3$. You can view this as $4 \cdot x^3$. The derivative is:

$h'(x) = 4 \cdot \frac{d}{dx}[x^3]$

You’ll compute $\frac{d}{dx}[x^3]$ using the power rule (next major topic), but the key idea here is that the $4$ stays out front.

Example 2

If $p(x) = -\frac{1}{2} \sin(x)$, then:

$p'(x) = -\frac{1}{2} \cos(x)$

The trig derivative comes later, but the constant multiple rule is what keeps the factor $-\frac{1}{2}$ unchanged.

Sum and Difference Rules

If a function is the sum (or difference) of two differentiable functions, its derivative is the sum (or difference) of their derivatives.

If $f(x)$ and $g(x)$ are differentiable:

$\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)$

$\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)$

Why this matters: most AP-style derivatives are combinations like polynomials plus trig plus exponentials. These rules let you separate the work into manageable pieces.

Example 1

Differentiate $F(x) = x^4 + 3x - \cos(x)$.

Use sum/difference to differentiate term-by-term:

• $\frac{d}{dx}[x^4]$ will use the power rule.
• $\frac{d}{dx}[3x]$ uses constant multiple and power rule.
• $\frac{d}{dx}[-\cos(x)]$ uses constant multiple and trig derivatives.

So you can write:

$F'(x) = \frac{d}{dx}[x^4] + \frac{d}{dx}[3x] - \frac{d}{dx}[\cos(x)]$

Then evaluate each derivative with the appropriate rule.

Example 2 (watch the minus sign)

Let $G(x) = 5x^2 - (2x^3 - 1)$.

A common mistake is to differentiate without first handling the subtraction correctly. You can simplify first:

$G(x) = 5x^2 - 2x^3 + 1$

Now the difference rule applies cleanly term-by-term.

What goes wrong most often

• Students sometimes think $\frac{d}{dx}[f(x) + g(x)] = f'(x) g'(x)$. That is false. Addition does not turn into multiplication.
• Dropping a negative sign is extremely common, especially with expressions like $-(2x^3 - 1)$.
• Confusing “constant rule” with “power rule” for terms like $x^0$. Since $x^0 = 1$ (for $x \neq 0$), its derivative is $0$ because it is constant.

Exam Focus

• Typical question patterns:
  • Differentiate a multi-term expression (polynomial plus trig/exp/log) by separating into terms.
  • Find $f'(a)$ for a given $a$ after computing $f'(x)$.
  • Use derivative notation correctly, switching between $f'(x)$ and $\frac{dy}{dx}$.
• Common mistakes:
  • Turning sums into products (for example, writing $\frac{d}{dx}[x^2 + \sin(x)] = 2x\cos(x)$).
  • Losing negatives when distributing subtraction through parentheses.
  • Treating a constant multiple as if it changes the derivative rule (it doesn’t, it just stays as a factor).

The Power Rule

The power rule is the workhorse for differentiating polynomials and many algebraic expressions. It tells you how to differentiate $x$ raised to a constant power.

What it says

For any real number $n$, the derivative of $x^n$ (where defined) is:

$\frac{d}{dx}[x^n] = n x^{n-1}$

Why it matters

Polynomials (like $3x^4 - 2x + 9$) are built entirely out of powers of $x$ and constants. Once you can differentiate $x^n$, the linearity rules let you differentiate an entire polynomial quickly. The power rule also extends to negative powers (rational functions written as powers) and fractional powers (roots written as exponents), which show up often in AP problems.

How it works (mechanism)

Conceptually, the power rule is consistent with the limit definition of the derivative and algebraic expansion patterns. You don’t need to re-derive it on the exam, but you should understand what it does:

• The old exponent $n$ becomes a coefficient.
• The new exponent becomes $n - 1$.

That “bring down the power, subtract one” pattern is easy to remember, but it must be applied carefully, especially when the base is not exactly $x$ (that situation requires the chain rule, which is a different topic).

Applying the power rule to common forms

Polynomials

A polynomial is a sum of terms like $a x^n$ where $a$ is a constant and $n$ is a nonnegative integer. Use constant multiple + power rule term-by-term.

Example 1

Differentiate:

$f(x) = 3x^5 - 7x^2 + 4x - 9$

Differentiate each term:

$\frac{d}{dx}[3x^5] = 3 \cdot 5x^4 = 15x^4$

$\frac{d}{dx}[-7x^2] = -7 \cdot 2x = -14x$

$\frac{d}{dx}[4x] = 4 \cdot 1 = 4$

$\frac{d}{dx}[-9] = 0$

So:

$f'(x) = 15x^4 - 14x + 4$

Negative powers (rewrite first)

Negative exponents often appear as fractions. Rewriting helps you avoid quotient rule when it’s unnecessary.

Example 2

Differentiate:

$g(x) = \frac{5}{x^3}$

Rewrite as:

$g(x) = 5x^{-3}$

Apply constant multiple and power rule:

$g'(x) = 5 \cdot (-3)x^{-4}$

So:

$g'(x) = -15x^{-4}$

You can rewrite back as a fraction:

$g'(x) = -\frac{15}{x^4}$

Fractional powers (roots)

A square root is a power of $\frac{1}{2}$, a cube root is a power of $\frac{1}{3}$, and so on.

Example 3

Differentiate:

$h(x) = \sqrt{x} = x^{\frac{1}{2}}$

Then:

$h'(x) = \frac{1}{2}x^{-\frac{1}{2}}$

Which can be rewritten as:

$h'(x) = \frac{1}{2\sqrt{x}}$

What goes wrong most often

• Applying the power rule to something that is not $x^n$, like $(3x + 1)^5$. The power rule alone does not work there; you’d need the chain rule.
• Forgetting that $\frac{d}{dx}[x] = 1$, which is the power rule with $n = 1$.
• Mishandling negative exponents when rewriting (for example, thinking $x^{-3} = \frac{1}{x^{-3}}$ instead of $x^{-3} = \frac{1}{x^3}$).

Exam Focus

• Typical question patterns:
  • Differentiate polynomials and evaluate the derivative at a point (for example, find $f'(2)$).
  • Differentiate expressions with roots or rational forms by rewriting into exponent form.
  • Identify where a derivative is undefined due to domain issues (for example, $\frac{1}{x^2}$ at $x = 0$).
• Common mistakes:
  • Using the power rule on a composite base like $(2x - 5)^4$ without chain rule.
  • Dropping the exponent subtraction step (writing $nx^n$ instead of $nx^{n-1}$).
  • Algebra errors when rewriting between radicals, fractions, and exponents.

Derivatives of Trigonometric Functions

Trigonometric functions model periodic behavior: oscillations, waves, circular motion, and repeating signals. In calculus, their derivatives matter because the derivative of a trig function is another trig function, which makes trig functions “closed” under differentiation. This is especially useful in motion problems (position, velocity, acceleration) and in modeling cycles.

The core trig derivative facts

For angles measured in radians (the standard in calculus), the fundamental derivatives are:

$\frac{d}{dx}[\sin(x)] = \cos(x)$

$\frac{d}{dx}[\cos(x)] = -\sin(x)$

From these, you also need the derivatives of the other four trig functions:

$\frac{d}{dx}[\tan(x)] = \sec^2(x)$

$\frac{d}{dx}[\sec(x)] = \sec(x)\tan(x)$

$\frac{d}{dx}[\csc(x)] = -\csc(x)\cot(x)$

$\frac{d}{dx}[\cot(x)] = -\csc^2(x)$

Why “radians” matters

These derivative rules are true in this clean form only when $x$ is in radians. If you use degrees, extra conversion factors appear. On the AP exam, you should assume trig derivatives use radians unless explicitly stated otherwise.

How to apply these rules

You usually combine trig derivatives with linearity (sum/difference and constant multiple). Later, you’ll combine them with chain rule and product/quotient rules; for now, focus on recognizing which trig function you have and applying the matching derivative.

Example 1

Differentiate:

$f(x) = 6\sin(x) - 3\cos(x)$

Apply constant multiple and sum/difference:

$f'(x) = 6\cos(x) - 3(-\sin(x))$

So:

$f'(x) = 6\cos(x) + 3\sin(x)$

Example 2

Differentiate:

$g(x) = \tan(x) + 2\sec(x)$

Then:

$g'(x) = \sec^2(x) + 2\sec(x)\tan(x)$

Common conceptual pitfalls

• Sign errors: The derivative of $\cos(x)$ is negative. Many students remember “sine goes to cosine” but forget the negative when “cosine goes to negative sine.”
• Mixing up $\sec^2(x)$ and $\sec(x)\tan(x)$: $\tan(x)$ differentiates to $\sec^2(x)$, but $\sec(x)$ differentiates to $\sec(x)\tan(x)$.
• Undefined points: Functions like $\tan(x)$ and $\sec(x)$ are undefined where $\cos(x) = 0$. Their derivatives inherit those domain restrictions.

Exam Focus

• Typical question patterns:
  • Differentiate expressions combining trig functions with polynomials (for example, $x^2\sin(x)$ later uses product rule).
  • Compute $f'(a)$ for a given value and possibly simplify using trig identities.
  • Interpret a trig derivative in a rate-of-change context (periodic motion).
• Common mistakes:
  • Forgetting the negative in $\frac{d}{dx}[\cos(x)] = -\sin(x)$.
  • Writing $\frac{d}{dx}[\tan(x)] = \sec(x)\tan(x)$ (wrong; that’s for $\sec(x)$).
  • Treating trig derivatives as if they require the power rule (they don’t; trig functions have their own rules).

Derivatives of Exponential and Logarithmic Functions

Exponential and logarithmic functions show up whenever growth changes proportionally to current size (population growth, compound interest, radioactive decay) or whenever you’re measuring multiplicative change additively (logs). In calculus, they’re crucial because many differential equations and growth/decay models depend on them, and because $e^x$ has the special property that its derivative is itself.

Exponential functions

Derivative of $e^x$

The natural exponential function $e^x$ is special because:

$\frac{d}{dx}[e^x] = e^x$

Why this matters: $e^x$ is the unique base for which the function equals its own rate of change. This is why $e$ appears constantly in continuous growth models.

Example 1

Differentiate:

$f(x) = 5e^x - e^x + 2$

Combine like terms or just differentiate term-by-term:

$f'(x) = 5e^x - e^x + 0$

So:

$f'(x) = 4e^x$

Derivative of $a^x$

For an exponential function with base $a$ (where $a > 0$ and $a \neq 1$):

$\frac{d}{dx}[a^x] = a^x \ln(a)$

This formula is worth understanding: differentiating $a^x$ pulls down a factor of $\ln(a)$.

Example 2

Differentiate:

$g(x) = 2^x$

Then:

$g'(x) = 2^x \ln(2)$

A common simplification mistake is trying to turn $\ln(2)$ into a decimal on a non-calculator exam. Usually you should leave it as $\ln(2)$.

Logarithmic functions

Logarithms undo exponentials. The natural log $\ln(x)$ is the inverse of $e^x$, and its derivative is one of the most frequently used facts in calculus.

Derivative of $\ln(x)$

For $x > 0$:

$\frac{d}{dx}[\ln(x)] = \frac{1}{x}$

Domain matters: $\ln(x)$ is defined only for positive $x$ (in the real-number setting of AP Calculus). So the derivative formula is stated with $x > 0$.

Example 1

Differentiate:

$h(x) = 3\ln(x) - \frac{1}{x}$

First rewrite $\frac{1}{x}$ as $x^{-1}$ if you want:

$\frac{d}{dx}\left[-\frac{1}{x}\right] = \frac{d}{dx}[-x^{-1}] = -(-1)x^{-2} = x^{-2}$

And:

$\frac{d}{dx}[3\ln(x)] = 3\cdot \frac{1}{x}$

So:

$h'(x) = \frac{3}{x} + \frac{1}{x^2}$

Derivative of $\log_a(x)$

If $\log_a(x)$ is log base $a$ (with $a > 0$ and $a \neq 1$), then for $x > 0$:

$\frac{d}{dx}[\log_a(x)] = \frac{1}{x\ln(a)}$

This matches the change-of-base relationship between $\log_a(x)$ and $\ln(x)$.

Example 2

Differentiate:

$p(x) = \log_3(x)$

Then:

$p'(x) = \frac{1}{x\ln(3)}$

Common conceptual pitfalls

• Confusing $\ln(x)$ with $\frac{1}{\ln(x)}$. The derivative of $\ln(x)$ is $\frac{1}{x}$, not something involving $\ln(x)$.
• Forgetting domain restrictions: $\ln(x)$ and $\log_a(x)$ require $x > 0$ in this course.
• Mixing up the exponential derivative rules: $\frac{d}{dx}[e^x] = e^x$ but $\frac{d}{dx}[a^x] = a^x\ln(a)$.

Exam Focus

• Typical question patterns:
  • Differentiate expressions containing $e^x$, $a^x$, $\ln(x)$, and simplify.
  • Evaluate derivatives at points (for example, find the slope of the tangent to $y = 2^x$ at $x = 0$).
  • Combine log/exp derivatives with product or quotient rules in more complex expressions.
• Common mistakes:
  • Writing $\frac{d}{dx}[\ln(x)] = \ln(x)$ (confusing it with $e^x$).
  • Omitting the $\ln(a)$ factor when differentiating $a^x$.
  • Ignoring that $\ln(x)$ is undefined for $x \le 0$ (which can affect whether an answer makes sense for a given interval).

The Product Rule

Real AP problems rarely hand you a function that is just a simple sum of basic pieces. Very often, you’ll see products like $x^2\sin(x)$ or $(x+1)e^x$. The product rule is the tool for differentiating the product of two differentiable functions.

What it is

If $h(x) = f(x)g(x)$, then:

$h'(x) = f'(x)g(x) + f(x)g'(x)$

Why it matters (and why you can’t just “multiply the derivatives”)

A tempting but incorrect idea is:

$\frac{d}{dx}[f(x)g(x)] = f'(x)g'(x)$

That is not generally true. The correct product rule reflects that both factors can change with $x$, and the rate of change of the product depends on two contributions:

• “First changes, second stays” gives $f'(x)g(x)$.
• “Second changes, first stays” gives $f(x)g'(x)$.

A helpful analogy: if the area of a rectangle is $A = LW$, then a small change in area comes from changing the length while width stays about the same, plus changing the width while length stays about the same.

How to use it step-by-step

1. Identify $f(x)$ and $g(x)$.
2. Compute $f'(x)$ and $g'(x)$ using basic rules.
3. Substitute into $f'(x)g(x) + f(x)g'(x)$.
4. Simplify if asked (but don’t overdo algebra if it risks mistakes).

Example 1

Differentiate:

$h(x) = x^2\sin(x)$

Let $f(x) = x^2$ and $g(x) = \sin(x)$.

Then:

$f'(x) = 2x$

$g'(x) = \cos(x)$

Apply product rule:

$h'(x) = (2x)\sin(x) + (x^2)\cos(x)$

So:

$h'(x) = 2x\sin(x) + x^2\cos(x)$

Example 2

Differentiate:

$p(x) = (x+1)e^x$

Let $f(x) = x+1$ and $g(x) = e^x$.

Then:

$f'(x) = 1$

$g'(x) = e^x$

So:

$p'(x) = (1)e^x + (x+1)e^x$

Combine like terms:

$p'(x) = (x+2)e^x$

What goes wrong most often

• Forgetting one of the two terms. The product rule always produces a sum of two products.
• Misidentifying $f(x)$ and $g(x)$ in a complicated expression, leading to wrong derivatives.
• Expanding unnecessarily: Sometimes expanding first helps (especially with polynomials), but expanding $x^2\sin(x)$ isn’t possible in a useful algebraic way. Choose the approach that reduces error.

Exam Focus

• Typical question patterns:
  • Differentiate a product of an algebraic term and a trig/exp/log term.
  • Find the equation of a tangent line to a product function at a given $x$ value.
  • Use product rule inside a larger expression with sums and constants.
• Common mistakes:
  • Using $f'(x)g'(x)$ instead of $f'(x)g(x) + f(x)g'(x)$.
  • Forgetting to multiply by the original (undifferentiated) factor in each term.
  • Algebra mistakes when simplifying, especially factoring out common terms like $e^x$.

The Quotient Rule

When a function is a ratio of two differentiable functions, you can sometimes rewrite it using negative exponents and use the power rule instead. But when rewriting isn’t convenient (or when the denominator is not a simple power of $x$), the quotient rule is the standard tool.

What it is

If:

$h(x) = \frac{f(x)}{g(x)}$

where $g(x) \neq 0$, then:

$h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}$

Why it matters

Quotients appear in rates (for example, “per” quantities), rational functions, and many applied contexts. The quotient rule helps you differentiate without doing long algebraic division or complicated rewrites.

A memory aid that many students use is:

• “Low d-high minus high d-low over low-low.”

Translated: denominator times derivative of numerator minus numerator times derivative of denominator, all over denominator squared.

How to use it step-by-step

1. Identify the numerator as $f(x)$ and denominator as $g(x)$.
2. Compute $f'(x)$ and $g'(x)$.
3. Substitute into the formula carefully, paying attention to parentheses.
4. Simplify cautiously.

Example 1

Differentiate:

$q(x) = \frac{x^2 + 1}{x}$

Here $f(x) = x^2 + 1$ and $g(x) = x$.

Compute derivatives:

$f'(x) = 2x$

$g'(x) = 1$

Apply quotient rule:

$q'(x) = \frac{(2x)(x) - (x^2 + 1)(1)}{x^2}$

Simplify numerator:

$q'(x) = \frac{2x^2 - x^2 - 1}{x^2}$

So:

$q'(x) = \frac{x^2 - 1}{x^2}$

Notice you could also rewrite $q(x) = x + x^{-1}$ and use power rule to get $q'(x) = 1 - x^{-2}$, which is equivalent to $\frac{x^2 - 1}{x^2}$. This is a good reminder: sometimes there are multiple correct methods.

Example 2

Differentiate:

$r(x) = \frac{\sin(x)}{x^2}$

Let $f(x) = \sin(x)$ and $g(x) = x^2$.

Then:

$f'(x) = \cos(x)$

$g'(x) = 2x$

Apply quotient rule:

$r'(x) = \frac{(\cos(x))(x^2) - (\sin(x))(2x)}{(x^2)^2}$

So:

$r'(x) = \frac{x^2\cos(x) - 2x\sin(x)}{x^4}$

You can factor $x$ in the numerator if you want, but it’s not required unless the problem asks for a specific form.

What goes wrong most often

• Sign error in the numerator: It is $f'(x)g(x) - f(x)g'(x)$, not a plus.
• Forgetting to square the denominator: The denominator becomes $(g(x))^2$.
• Dropping parentheses: If $f(x)$ or $g(x)$ is multi-term, you must keep it grouped when substituting.

Exam Focus

• Typical question patterns:
  • Differentiate a rational function where rewriting is possible but not obvious, testing your strategic choice.
  • Compute a slope at a point for a quotient function, sometimes with trig in the numerator or denominator.
  • Combine quotient rule with simplification and domain awareness (where $g(x) = 0$ the function and derivative are undefined).
• Common mistakes:
  • Writing the numerator as $f'(x)g'(x) - f(x)g(x)$ (mixing up the pattern).
  • Squaring the wrong expression (for example, squaring only part of the denominator).
  • Cancelling factors incorrectly across subtraction in the numerator (you can only factor and cancel a common factor, not “cancel terms”).