Unit 2: Differentiation: Definition and Fundamental Properties

Defining the Derivative

Derivative at a Point

• The derivative of a function at a specific point \(x = a\) measures the instantaneous rate of change of the function at that point. It is formally defined as: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \] This limit represents the slope of the tangent line to the graph at \(x = a\). The concept extends the idea of average rate of change over an interval to the case where the interval length approaches zero. On the AP exam, this definition is often used to justify derivative values without applying shortcut rules.
• The difference quotient \(\frac{f(a+h) - f(a)}{h}\) gives the slope of a secant line between the points \((a, f(a))\) and \((a+h, f(a+h))\). As \(h\) approaches zero, the secant line becomes the tangent line, and the slope approaches the derivative. This process captures the idea of an instantaneous slope, something not possible with algebra alone. This is why limits are a prerequisite for understanding derivatives. It’s also why Unit 1 and Unit 2 are deeply connected.
• Derivatives at a point can also be defined using an alternative limit form: \[ f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} \] This form approaches \(a\) directly instead of shifting by \(h\). Both definitions are equivalent and interchangeable, but one form may be easier depending on the problem. Recognizing both is useful for AP questions that present derivatives in unfamiliar forms.
• When a function’s derivative at a point exists, the function is said to be differentiable at that point. Differentiability implies that the graph is locally linear near that point—meaning it can be well-approximated by its tangent line. If the derivative does not exist, the function might have a corner, cusp, vertical tangent, or discontinuity at that point. Knowing these conditions is essential for analyzing motion, optimization, and graphing problems later in the course.
• In real-world applications, derivatives at a point often represent quantities like velocity, acceleration, or growth rate at a specific time or condition. For example, in physics, the derivative of position with respect to time gives instantaneous velocity. In economics, the derivative of cost with respect to production quantity gives marginal cost. This makes the concept of “derivative at a point” both mathematically rigorous and practically meaningful.

Derivative as a Function

• The derivative as a function, written \(f'(x)\), gives the slope of the tangent line for every value of \(x\) in the domain where the derivative exists. This creates a new function derived from the original one. For example, if \(f(x) = x^2\), then \(f'(x) = 2x\), meaning the slope changes depending on \(x\). This perspective allows us to analyze trends in slope across an entire graph instead of just at one point.
• To find the derivative function from the definition, we replace \(a\) with \(x\) in the limit

  Defining the Derivative

  Derivative of a Function at a Point and as a Function

  • The derivative of a function at a point \(x = a\) is defined as \(\lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\), provided this limit exists. This definition captures the instantaneous rate of change of the function at \(x = a\). In a geometric sense, it represents the slope of the tangent line to the graph at that point. If the limit does not exist or is infinite, the function is not differentiable at that point. This concept is foundational because every derivative rule stems from this definition.
  • The derivative of a function as a function, \(f'(x)\), is obtained by replacing \(a\) with \(x\) in the limit definition: \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). This provides a formula that works for any \(x\) in the function’s domain where the limit exists. Rather than finding the slope at a single point, this process finds a general expression for the slope at any point. This is what allows calculus to handle dynamic situations, where rates of change vary continuously. On the AP exam, both forms of the definition are tested.
  • There is an alternate form of the derivative definition: \(\lim_{x \to a} \frac{f(x) - f(a)}{x-a}\). This approach is sometimes easier when given a specific \(a\) because it avoids introducing \(h\). Both definitions are mathematically equivalent and measure the same concept—the slope of the tangent line at \(x = a\). Recognizing both forms helps when solving limits that are easier in one format over the other. AP problems may require switching between these forms fluently.
  • The derivative at a point can be interpreted as the limit of the slopes of secant lines connecting \((a, f(a))\) to nearby points on the curve. As the second point gets infinitely close to \(a\), the secant line approaches the tangent line. This visualization links algebraic definitions to geometric intuition. Understanding this link helps in interpreting graphs, especially in free-response questions where reasoning is required.
  • The process of finding a derivative from the limit definition is called differentiation. While the limit definition is used for proofs and conceptual understanding, in practice we rely on differentiation rules for efficiency. However, the AP exam often includes at least one question requiring evaluation from the definition to test understanding of the concept. Mastery of the definition is critical because it is the theoretical foundation for every differentiation shortcut.

  Interpreting the Derivative

  Graphical, Physical, and Contextual Meaning

  • Derivative as slope of the tangent line: At a point \(x=a\), \(f'(a)\) is the slope of the tangent line to \(y=f(x)\). A positive derivative means the graph is rising at that point; a negative derivative means it is falling. A derivative of zero indicates a horizontal tangent, which may be a local max, local min, or neither (e.g., a saddle). Interpreting \(f'(a)\) this way connects algebraic limits to the geometry of curves and the behavior of functions near \(x=a\).
  • Instantaneous rate of change with units: The derivative carries units of “output per input,” so if \(f\) is position in meters and \(x\) is time in seconds, \(f'(t)\) is velocity in m/s. Magnitude tells how fast the output is changing, while sign tells the direction of change. Including units in interpretations earns credit on AP free-response items and prevents misstatements. Always tie your conclusion to the real-world quantity asked for, not just “increase/decrease.”
  • Reading \(f'(x)\) from a graph or table: On a graph of \(y=f(x)\), estimate \(f'(a)\) by drawing a good tangent and computing rise/run from two nearby points. If you are given a table of \(f(x)\), approximate \(f'(a)\) with a symmetric difference quotient, \(\dfrac{f(a+h)-f(a-h)}{2h}\), using the closest available values. Recognize that larger step sizes make the approximation less accurate. Clearly state your method and show numerical work to receive full credit.
  • Motion interpretation (position–velocity–acceleration): If \(s(t)\) is position, then \(s'(t)=v(t)\) is velocity and \(s''(t)=a(t)\) is acceleration. When \(v(t)>0\) the object moves in the positive direction; when \(v(t)<0\) it moves in the negative direction. If \(a(t)\) and \(v(t)\) have the same sign, speed increases; if they have opposite signs, speed decreases. These sign charts are a fast way to answer qualitative questions about motion on the AP exam.
  • Average vs. instantaneous change: Average rate on \([a,b]\) is the slope of the secant, \(\dfrac{f(b)-f(a)}{b-a}\), while instantaneous rate at \(a\) is \(f'(a)\), the limit of those secant slopes. If the average and instantaneous rates differ, the function must curve between \(a\) and \(b\). Recognizing this distinction avoids common errors like substituting an average when a problem asks for instantaneous change. Always check the prompt’s wording: “at \(x=a\)” signals a derivative, not an average.
  • Local linearity and tangent-line models: Near \(x=a\), a differentiable function behaves approximately like its tangent line \(L(x)=f(a)+f'(a)(x-a)\). This linear model translates slope information into quick estimates of function values without a calculator. The accuracy improves as \(x\) gets closer to \(a\), reflecting the limit idea behind the derivative. On AP problems, this interpretation often appears as “use the tangent line to approximate \(f(a+\Delta x)\).”

  Connecting Differentiability and Continuity

  Relationship and Conditions

  • Differentiability implies continuity: If a function is differentiable at \(x=a\), then it must be continuous at \(x=a\). This follows because the derivative’s limit definition requires the function to approach a single value at \(a\) from both sides. However, the reverse is not true—continuity does not guarantee differentiability. This logical one-way relationship is important for AP justifications when explaining why a function must be continuous before certain derivative properties apply.
  • Continuity without differentiability: A function can be continuous at a point but not differentiable there, typically due to sharp corners, cusps, or vertical tangent lines. For example, \(f(x) = |x|\) is continuous everywhere but not differentiable at \(x=0\) because the slope changes abruptly from negative to positive. Similarly, \(f(x) = x^{1/3}\) has a vertical tangent at \(x=0\), where the derivative approaches infinity. Recognizing these cases prevents overgeneralizing continuity into differentiability.
  • Algebraic test for differentiability: For piecewise functions, differentiability at a boundary point requires both continuity and equal one-sided derivatives at that point. First, confirm the left- and right-hand limits of the function match the function’s value. Then, compute the derivatives of each piece and check they match at the boundary. If the slopes differ, the function has a corner and is not differentiable, even if it’s continuous there.
  • Graphical clues for non-differentiability: In addition to corners and vertical tangents, oscillations that become infinitely frequent near a point can break differentiability. For example, \(f(x) = x\sin(\frac{1}{x})\) is continuous at \(x=0\) but has a derivative that fails to exist there due to rapid oscillations in slope. On the AP exam, graphs may show subtle cases like this, so watch for erratic slope behavior. Identifying these ensures correct application of derivative-based theorems.
  • Implications for applied problems: In real-world contexts, non-differentiable points often correspond to abrupt changes in physical conditions—such as a car suddenly reversing direction or a piecewise cost function changing rates. This makes differentiability checks crucial when modeling smooth vs. abrupt changes. In AP free-response questions, explicitly stating whether a function is differentiable (and why) can be part of earning justification points for conclusions about motion, optimization, or curve sketching.

  Determining Derivatives for Elementary Functions

  Polynomials, Exponentials, and Trigonometric Functions

  • Polynomials: For \(f(x) = x^n\) where \(n\) is a positive integer, the derivative is given by \(f'(x) = nx^{n-1}\). This is known as the power rule and can be derived directly from the limit definition. For example, \( \frac{d}{dx}(x^3) = 3x^2 \) because the slopes of secant lines approach this value as \(h \to 0\). The power rule extends to sums and differences of polynomial terms using the sum/difference rule. This efficiency is why we rarely compute polynomial derivatives from scratch using limits in practice.
  • Exponential functions: For \(f(x) = e^x\), the derivative is \(f'(x) = e^x\), meaning it is its own rate of change. This property comes from the special base \(e\) in which the average rate of change approaches the same value as the function itself. More generally, for \(a^x\) with \(a > 0\), \( \frac{d}{dx}(a^x) = a^x \ln(a) \). This constant multiple \(\ln(a)\) emerges from applying the chain rule to the natural exponential function.
  • Trigonometric functions: Using the limit definition and the Squeeze Theorem, we find \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\). These results come from the two critical trig limits: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\lim_{x \to 0} \frac{\cos x - 1}{x} = 0\). These derivatives form the foundation for all other trig derivative rules, many of which can be derived using the quotient and product rules. On the AP exam, these appear frequently in both symbolic and applied contexts.
  • Other basic trig derivatives: From the quotient identities, we get \(\frac{d}{dx}(\tan x) = \sec^2 x\), \(\frac{d}{dx}(\cot x) = -\csc^2 x\), \(\frac{d}{dx}(\sec x) = \sec x \tan x\), and \(\frac{d}{dx}(\csc x) = -\csc x \cot x\). Each has a restricted domain where it is differentiable, excluding points where the function is undefined (such as \(x = \frac{\pi}{2} + n\pi\) for \(\tan x\)). Recognizing these domains is important when determining differentiability in trig-based problems. AP questions often test whether students recall both the formulas and their restrictions.
  • Verifying formulas from the limit definition: While differentiation rules save time, AP problems may require proving a derivative formula using \(\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\). For example, proving \(\frac{d}{dx}(\sin x) = \cos x\) requires applying trig identities and the two fundamental trig limits. This reinforces that every shortcut in differentiation stems from the fundamental limit definition. Being able to work from both perspectives ensures flexibility on conceptual questions.

  Basic Differentiation Rules

  Constant, Constant Multiple, and Sum/Difference Rules

  • Constant Rule: The derivative of a constant \(k\) is always zero, \(\frac{d}{dx}(k) = 0\), because constants do not change and thus have no rate of change. For example, \(\frac{d}{dx}(7) = 0\) and \(\frac{d}{dx}(-\pi) = 0\). Graphically, the slope of a horizontal line is zero everywhere, which directly reflects this rule. This rule may seem simple but appears often in algebraic simplification during complex differentiation problems.
  • Constant Multiple Rule: If \(c\) is a constant and \(f\) is differentiable, then \(\frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)\). This means we can factor constants out before differentiating, making problems cleaner and faster to solve. For example, \(\frac{d}{dx}(5x^3) = 5(3x^2) = 15x^2\). Recognizing when to apply this saves time, especially in AP multiple-choice questions where simplification speed matters.
  • Sum Rule: If \(f\) and \(g\) are differentiable, then \(\frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x)\). The derivative of a sum is simply the sum of the derivatives, allowing term-by-term differentiation. This is particularly useful for polynomials, where each term can be handled independently. For example, \(\frac{d}{dx}(x^3 + 4x - 7) = 3x^2 + 4\).
  • Difference Rule: Similar to the sum rule, \(\frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x)\). The derivative distributes over subtraction exactly as it does over addition, but the minus sign must be preserved. For example, \(\frac{d}{dx}(x^5 - 2x^2) = 5x^4 - 4x\). Forgetting to distribute the negative sign when differentiating subtraction is a common AP exam mistake.
  • Combining basic rules efficiently: Most simple functions can be differentiated quickly by combining these rules without writing every step. For example, \(\frac{d}{dx}(3x^4 - 5x + 7) = 12x^3 - 5\) is found by applying the constant multiple rule and sum/difference rule simultaneously. While these rules are straightforward, they form the building blocks for more advanced differentiation techniques like the product, quotient, and chain rules. Strong fluency here speeds up work on more complicated derivatives later.

  Derivatives of Trigonometric Functions

  Formulas, Derivations, and Domain Considerations

  • Derivatives of sine and cosine: Using the limit definition of the derivative along with the special trig limits \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) and \(\lim_{x \to 0} \frac{\cos x - 1}{x} = 0\), we find \(\frac{d}{dx}(\sin x) = \cos x\) and \(\frac{d}{dx}(\cos x) = -\sin x\). These results are fundamental and appear frequently in AP Calculus problems, both in symbolic differentiation and in motion modeling. Because sine and cosine are continuous and differentiable everywhere, their derivatives are also defined for all real \(x\).
  • Derivatives of tangent and cotangent: By differentiating \(\tan x = \frac{\sin x}{\cos x}\) using the quotient rule, we get \(\frac{d}{dx}(\tan x) = \sec^2 x\). Similarly, \(\frac{d}{dx}(\cot x) = -\csc^2 x\). Both tangent and cotangent have vertical asymptotes where their respective denominators (\(\cos x\) for \(\tan x\), \(\sin x\) for \(\cot x\)) are zero, so their derivatives are undefined at those points. Recognizing these domain restrictions is important when discussing differentiability.
  • Derivatives of secant and cosecant: Differentiating \(\sec x = \frac{1}{\cos x}\) and \(\csc x = \frac{1}{\sin x}\) via the quotient rule gives \(\frac{d}{dx}(\sec x) = \sec x \tan x\) and \(\frac{d}{dx}(\csc x) = -\csc x \cot x\). These derivatives combine the original trig function with another trig function, which makes them easy to misremember. Writing them down systematically during test preparation helps avoid sign and pairing errors. These functions are undefined wherever their original denominators are zero, matching the derivative’s domain.
  • Pattern recognition and symmetry: The derivatives of sine and cosine cycle every four differentiations: \(\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x\). Tangent and secant derivatives involve positive signs, while cotangent and cosecant derivatives involve negatives. Remembering these sign patterns can speed up recall during timed sections of the AP exam. This cyclical property also reflects the periodic nature of trigonometric functions.
  • Applied interpretation of trig derivatives: In motion problems where position is given as a sine or cosine function, the derivative represents velocity, and the second derivative represents acceleration. For example, \(s(t) = \sin t\) means \(v(t) = \cos t\) and \(a(t) = -\sin t\), showing a phase shift in motion. These interpretations link the abstract derivative formulas to physical behavior in oscillating systems, a connection that often appears in AP free-response questions.

  Applying Differentiation Rules

  Product Rule, Quotient Rule, Chain Rule, and Higher-Order Derivatives

  • Product Rule: When differentiating the product of two functions \(f(x)\) and \(g(x)\), the rule is \((fg)' = f'(x)g(x) + f(x)g'(x)\). This means you differentiate one factor while keeping the other unchanged, then reverse the roles, and finally add the two results. A common mistake is trying to simply multiply the derivatives, which is incorrect. For example, \(\frac{d}{dx}[(x^2)(\sin x)] = 2x\sin x + x^2\cos x\). This rule is heavily tested in AP problems involving both algebraic and applied contexts.
  • Quotient Rule: For \(\frac{f(x)}{g(x)}\), where \(g(x) \neq 0\), the rule is \(\left(\frac{f}{g}\right)' = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}\). The order in the numerator matters (i.e. switching terms changes the sign). For example, \(\frac{d}{dx}\left(\frac{\ln x}{x^2}\right) = \frac{x^2 \cdot \frac{1}{x} - \ln x \cdot 2x}{x^4} = \frac{x - 2x\ln x}{x^4}\). Students often memorize this with the phrase “low d-high minus high d-low over low squared” to remember the correct order.
  • Chain Rule: When differentiating a composite function \(y = f(g(x))\), the rule is \(\frac{dy}{dx} = f'(g(x)) \cdot g'(x)\). This accounts for the fact that changes in \(x\) affect \(y\) indirectly through \(g(x)\). For example, \(\frac{d}{dx}[\sin(x^2)] = \cos(x^2) \cdot 2x\). Forgetting to multiply by the derivative of the “inside” function is a frequent AP exam error, so always identify the inner and outer functions before differentiating.
  • Combining multiple rules: Many functions require a mix of product, quotient, and chain rules in a single differentiation. For example, \(\frac{d}{dx}\left[ x^2 \cdot e^{3x} \sin x \right]\) requires the product rule applied twice, along with the chain rule for \(e^{3x}\). Planning the order of application before differentiating reduces algebraic mistakes. On the AP exam, such problems are common in both symbolic differentiation and applied rate-of-change scenarios.
  • Higher-order derivatives: The derivative of a derivative is called a second derivative, denoted \(f''(x)\) or \(\frac{d^2y}{dx^2}\). These represent rates of change of rates of change, such as acceleration from velocity. Higher-order derivatives (third, fourth, etc.) follow the same rules as first derivatives but are applied repeatedly. Recognizing their notation and meaning is important for interpreting motion and concavity in later units.

  Implicit Differentiation

  Differentiating Equations Where \(y\) is Not Isolated

  • Concept: Implicit differentiation is used when \(y\) is defined indirectly by an equation involving both \(x\) and \(y\), rather than being isolated as \(y = f(x)\). By differentiating both sides of the equation with respect to \(x\), we treat \(y\) as a function of \(x\) and apply the chain rule to any derivative involving \(y\). This allows us to find \(\frac{dy}{dx}\) without explicitly solving for \(y\) first. It is particularly useful for curves that are difficult or impossible to write in explicit form.
  • Chain rule on \(y\) terms: Whenever you differentiate a term containing \(y\), you must multiply by \(\frac{dy}{dx}\) because \(y\) changes as \(x\) changes. For example, differentiating \(y^3\) with respect to \(x\) yields \(3y^2 \frac{dy}{dx}\). Forgetting to include \(\frac{dy}{dx}\) is the most common mistake in implicit differentiation problems. Keeping this rule in mind ensures correct application of the chain rule throughout the equation.
  • Solving for \(\frac{dy}{dx}\): After differentiating both sides, collect all terms containing \(\frac{dy}{dx}\) on one side and factor it out. This step allows you to isolate \(\frac{dy}{dx}\) and express it in terms of \(x\) and \(y\). For example, from \(2x + 3y^2 \frac{dy}{dx} = 0\), you get \(\frac{dy}{dx} = \frac{-2x}{3y^2}\). This final formula can then be used to compute slopes at specific points on the curve.
  • Example: Given \(x^2 + y^2 = 25\), differentiating both sides yields \(2x + 2y \frac{dy}{dx} = 0\). Solving for \(\frac{dy}{dx}\) gives \(\frac{dy}{dx} = -\frac{x}{y}\). This formula describes the slope of the tangent line to the circle of radius 5 at any point \((x, y)\). Such geometric applications frequently appear on the AP exam, often requiring substitution of a specific point to get a numerical slope.
  • Advanced applications: Implicit differentiation is essential for finding derivatives of inverse functions, logarithmic derivatives, and related rates problems. It also appears in multivariable calculus, where partial derivatives are taken with respect to one variable while treating others as dependent. Mastery of implicit differentiation ensures you can handle derivative problems even when the function’s explicit form is inaccessible.

  Example Problem 1: Derivative from the Definition

  Problem

  Use the limit definition to find the derivative of \(f(x) = 3x^2 - 5x\) at \(x = 2\).

  Step-by-Step Solution

  Step 1: Recall the definition: \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\).

  Step 2: Substitute \(a = 2\): \[ f'(2) = \lim_{h \to 0} \frac{[3(2+h)^2 - 5(2+h)] - [3(4) - 5(2)]}{h} \]

  Step 3: Expand and simplify: \[ = \lim_{h \to 0} \frac{[3(4+4h+h^2) - 10 - 5h] - [12 - 10]}{h} \] \[ = \lim_{h \to 0} \frac{(12+12h+3h^2 - 10 - 5h) - 2}{h} \]

  Step 4: Simplify numerator: \[ 12+12h+3h^2 - 10 - 5h - 2 = 7h + 3h^2 \] So: \[ f'(2) = \lim_{h \to 0} \frac{7h + 3h^2}{h} \]

  Step 5: Cancel \(h\): \[ f'(2) = \lim_{h \to 0} (7 + 3h) = 7 \]

  Final Answer: The derivative at \(x = 2\) is \(f'(2) = 7\).

  Example Problem 2: Applying Product and Chain Rules

  Problem

  Find the derivative of \(y = x^2 \cdot e^{3x}\).

  Step-by-Step Solution

  Step 1: Identify that this is a product of \(f(x) = x^2\) and \(g(x) = e^{3x}\), requiring the product rule: \((fg)' = f'(x)g(x) + f(x)g'(x)\).

  Step 2: Differentiate \(f(x) = x^2\) to get \(f'(x) = 2x\).

  Step 3: Differentiate \(g(x) = e^{3x}\) using the chain rule: \(g'(x) = 3e^{3x}\).

  Step 4: Apply the product rule: \[ y' = (2x)(e^{3x}) + (x^2)(3e^{3x}) \]

  Step 5: Factor if desired: \[ y' = e^{3x}(2x + 3x^2) \] This cleaner form often makes later work easier.

Final Answer: The derivative is \(y' = e^{3x}(2x + 3x^2)\).

Common Misconceptions

Frequent Errors and How to Avoid Them

1. Forgetting the chain rule on inner functions: A common mistake is differentiating the outer function but ignoring the derivative of the inside. For example, \(\frac{d}{dx}[\sin(5x)]\) is \(\cos(5x) \cdot 5\), not just \(\cos(5x)\). Always identify inner and outer functions before differentiating.

2. Mixing up product and quotient rules: Students sometimes apply the product rule to quotients or the quotient rule to products, leading to incorrect signs or terms. Remember: quotient rule involves subtraction in the numerator and a squared denominator, while the product rule is symmetrical and additive. Writing the rule before substituting can help prevent mistakes.

3. Dropping terms when applying the product rule: Forgetting one of the two required terms (differentiate first, keep second + keep first, differentiate second) is a common error. This omission often occurs under time pressure on the AP exam. Double-check by counting terms before moving on.

4. Assuming differentiability without checking continuity: Differentiability implies continuity, but the reverse is not always true. Functions can be continuous but not differentiable at corners, cusps, or vertical tangents. Always check slope behavior in addition to continuity when justifying differentiability.

5. Misremembering trigonometric derivatives: Swapping \(\sin\) and \(\cos\) derivatives or forgetting signs for \(\cos\), \(\cot\), and \(\csc\) is a frequent source of lost points. Writing all six trig derivatives at the start of the test can help avoid this mistake. Pay special attention to domain restrictions when discussing differentiability.

6. Not simplifying before differentiating: While rules can be applied to any form, simplifying algebraically first often makes the process shorter and reduces errors. For example, expanding \((x+1)^2\) before differentiating avoids a chain rule step. On the AP exam, strategic simplification can save time and help catch algebra slips.