Differentiation | Free AP Calculus AB Course

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<h1>AP Calculus AB: Differentiation</h1>

Differentiation is the engine of AP Calculus AB: nearly every later topic, from curve sketching to related rates to motion problems, depends on being fluent with these rules. This free, complete lesson covers the definition of the derivative, the Product, Quotient, and Chain Rules, implicit differentiation, derivatives of inverse functions, differentiability and continuity (including estimating derivatives numerically and graphically), the Mean Value Theorem, and L'Hospital's Rule. Each idea is followed by a fully worked example, and then a set of original practice problems with instant feedback and full explanations. Everything here is free, and you can keep practicing afterward with the full AP Calculus AB Question Bank linked below.

</p>

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<h2>What's covered in this lesson</h2>

<ol>

<li><a href="#definition">The Definition of the Derivative</a></li>

<li><a href="#rules">The Product, Quotient, and Chain Rules</a></li>

<li><a href="#implicit">Implicit Differentiation</a></li>

<li><a href="#inverse">Derivatives of Inverse Functions</a></li>

<li><a href="#diff-cont">Differentiability, Continuity, and Estimating Derivatives</a></li>

<li><a href="#mvt">The Mean Value Theorem</a></li>

<li><a href="#lhopital">L'Hospital's Rule and Recognizing a Limit as a Derivative</a></li>

<li><a href="#mistakes">Common Mistakes to Avoid</a></li>

<li><a href="#faq">Frequently Asked Questions</a></li>

</ol>

</nav>

<h2 id="definition">1. The Definition of the Derivative</h2>

<p>At any x in its domain, the <strong>derivative</strong> of y = f(x) is defined as the limit of the difference quotient:</p>

<p>Geometrically, the difference quotient is the slope of a secant line; its limit is the slope of the tangent line, the <strong>instantaneous rate of change</strong> of f at that point. The <strong>second derivative</strong>, f''(x), is simply the derivative of f'(x).</p>

<p><strong>Worked Example:</strong> Use the limit definition to find f'(x) for f(x) = x² + 3x.</p>

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<p class="prompt">1. Use the limit definition to find f'(x) for f(x) = x² + 5x.</p>

</div>

<p><span class="label">Explanation:</span> [(x+h)² + 5(x+h) − (x² + 5x)]/h = (2xh + h² + 5h)/h = 2x + h + 5, which approaches 2x + 5 as h → 0.</p>

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<p class="prompt">2. For f(x) = 4x − 7, find f'(3) using the difference quotient [f(3 + h) − f(3)]/h as h → 0.</p>

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<p><span class="label">Explanation:</span> f(3+h) = 5 + 4h and f(3) = 5, so the difference quotient is 4h/h = 4 for every h ≠ 0, giving a limit of 4.</p>

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<p class="prompt">3. As h → 0, a function's difference quotient [f(a + h) − f(a)]/h approaches 12. What does this represent geometrically?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p1-3',true)">A) The slope of the tangent line to the curve at x = a</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">B) The y-intercept of the curve</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">C) The area under the curve</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-3',false)">D) The value of f(a)</button>

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<p><span class="label">Explanation:</span> The limit of the difference quotient is, by definition, f'(a), the slope of the tangent line at that point.</p>

</div>

<p class="prompt">4. The second derivative f''(x) is defined as…</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p1-4',true)">A) The derivative of f'(x)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">B) The derivative of f(x) squared</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">C) f(x) divided by x</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p1-4',false)">D) The inverse of f'(x)</button>

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<p><span class="label">Explanation:</span> The second derivative is just the first derivative of f'(x), applying the same limit definition a second time.</p>

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<h2 id="rules">2. The Product, Quotient, and Chain Rules</h2>

<p>The <strong>Chain Rule</strong> differentiates a composite function: find the derivative of the "outside" function first, then multiply by the derivative of the "inside" one. Formally, if y = f(u) and u = g(x), then <span class="frac"><span class="num">dy</span><span class="den">dx</span></span> = <span class="frac"><span class="num">dy</span><span class="den">du</span></span> · <span class="frac"><span class="num">du</span><span class="den">dx</span></span>.</p>

<p><strong>Worked Example:</strong> If y = (x² + 3x − 1)⁴, find <span class="frac"><span class="num">dy</span><span class="den">dx</span></span> using the Chain Rule.</p>

</div>

</div>

<p><span class="label">Explanation:</span> By the Chain Rule: 5(x³ + 4)⁴ · 3x² = 15x²(x³ + 4)⁴.</p>

</div>

</div>

<p><span class="label">Explanation:</span> Writing as (5x − 3)<sup>1/2</sup>: <span class="frac"><span class="num">1</span><span class="den">2</span></span>(5x − 3)<sup>−1/2</sup> · 5 = <span class="frac"><span class="num">5</span><span class="den">2√(5x − 3)</span></span>.</p>

</div>

<p class="prompt">3. If f(t) = (t² + 1)sin(2t), find f'(0) using the Product Rule.</p>

</div>

<p><span class="label">Explanation:</span> f'(t) = (t²+1) · 2cos(2t) + sin(2t) · 2t. At t = 0: (1)(2)(1) + (0)(0) = 2.</p>

</div>

<p class="prompt">4. If f(v) = <span class="frac"><span class="num">3v</span><span class="den">1 − v²</span></span>, find f'(v) using the Quotient Rule, as a single fraction.</p>

</div>

<p><span class="label">Explanation:</span> f'(v) = <span class="frac"><span class="num">(1 − v²)(3) − 3v(−2v)</span><span class="den">(1 − v²)²</span></span> = <span class="frac"><span class="num">3 − 3v² + 6v²</span><span class="den">(1 − v²)²</span></span> = <span class="frac"><span class="num">3 + 3v²</span><span class="den">(1 − v²)²</span></span>.</p>

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</div>

<p><span class="label">Explanation:</span> By the Chain Rule: cos(3x² + 1) · 6x.</p>

</div>

</div>

<p><span class="label">Explanation:</span> By the Chain Rule: e<sup>cos x</sup> · (−sin x) = −sin(x) · e<sup>cos x</sup>.</p>

</div>

<h2 id="implicit">3. Implicit Differentiation</h2>

<p>When y is defined implicitly by an equation F(x, y) = 0 rather than solved explicitly, differentiate both sides with respect to x (treating y as a differentiable function of x and applying the Chain Rule to any y-term), then solve for <span class="frac"><span class="num">dy</span><span class="den">dx</span></span>.</p>

<p><strong>Worked Example:</strong> If x² + y² = 25, find <span class="frac"><span class="num">dy</span><span class="den">dx</span></span>.</p>

</div>

</div>

<p><span class="label">Explanation:</span> 2x − 3[x(dy/dx) + y] + 2y(dy/dx) = 0, so (dy/dx)(2y − 3x) = 3y − 2x, giving <span class="frac"><span class="num">3y − 2x</span><span class="den">2y − 3x</span></span>.</p>

</div>

</div>

<p><span class="label">Explanation:</span> 3x² + 3y²(dy/dx) = 6[x(dy/dx) + y], so (dy/dx)(3y² − 6x) = 6y − 3x². Dividing by 3: <span class="frac"><span class="num">2y − x²</span><span class="den">y² − 2x</span></span>.</p>

</div>

<p class="prompt">3. For the curve x² + 4y² = 16, at what point(s) is the tangent line vertical?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p3-3',false)">D) There are no such points</button>

</div>

<p><span class="label">Explanation:</span> Implicitly, dy/dx = −x/(4y), which is undefined (vertical tangent) when y = 0. Substituting y = 0 into the original equation: x² = 16, so x = ±4.</p>

</div>

<p class="prompt">4. Using implicit differentiation on x = cos y, verify the derivative of y = cos⁻¹x (arccos x).</p>

</div>

<p><span class="label">Explanation:</span> Differentiating x = cos y: 1 = −sin y(dy/dx), so dy/dx = −1/sin y. Since sin y ≥ 0 for 0 ≤ y ≤ π, sin y = √(1 − x²), giving <span class="frac"><span class="num">−1</span><span class="den">√(1 − x²)</span></span>.</p>

</div>

<h2 id="inverse">4. Derivatives of Inverse Functions</h2>

<p>If f and g are inverse functions, their graphs are reflections of each other across y = x. The derivative of the inverse of a function at a point is the <strong>reciprocal</strong> of the derivative of the function at the corresponding (swapped-coordinate) point:</p>

<p><strong>Worked Example:</strong> If f(4) = 10 and f'(4) = 6, what do we know about f⁻¹?</p>

<p>Since f passes through (4, 10), f⁻¹ must pass through (10, 4). And since f has slope 6 at that point, f⁻¹ must have slope <span class="frac"><span class="num">1</span><span class="den">6</span></span> at (10, 4).</p>

</div>

<p class="prompt">1. If f(2) = 9 and f'(2) = 3, what is the slope of f⁻¹ at x = 9?</p>

</div>

<p><span class="label">Explanation:</span> The slope of the inverse is the reciprocal: 1/f'(2) = 1/3.</p>

</div>

<p class="prompt">2. A function f satisfies f(1) = 5 with f'(1) = 2, and f(5) = 11 with f'(5) = 4. If g is the inverse of f, find g'(5).</p>

</div>

<p><span class="label">Explanation:</span> To find g'(5), look at the point on f where y = 5, namely (1, 5). Since f'(1) = 2, g'(5) = 1/2.</p>

</div>

<p class="prompt">3. Let y = f(x) = x³ + 2x − 3, and let g be the inverse function. Evaluate g'(0).</p>

</div>

<p><span class="label">Explanation:</span> f'(x) = 3x² + 2. By inspection, f(1) = 1 + 2 − 3 = 0, so x = 1 corresponds to y = 0. g'(0) = 1/f'(1) = 1/(3 + 2) = 1/5.</p>

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<p class="prompt">4. If f is one-to-one, differentiable, and f'(x) ≠ 0 everywhere, what is the general formula for (f⁻¹)'(x)?</p>

</div>

<p><span class="label">Explanation:</span> This is the general inverse-derivative formula: the reciprocal of f' evaluated at the corresponding point, f⁻¹(x).</p>

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<h2 id="diff-cont">5. Differentiability, Continuity, and Estimating Derivatives</h2>

<p>If f is differentiable at x = c, then f is automatically continuous there, this follows directly from the definition of the derivative. The converse is false: a function can be continuous at a point without being differentiable there, for example, at a corner, a cusp, or a vertical tangent. When only tabular data is available, the <strong>symmetric difference quotient</strong> gives a numerical estimate:</p>

<p class="prompt">1. If a function f is differentiable at x = c, what can we conclude?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-1',true)">A) f is continuous at x = c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-1',false)">B) f has a maximum at x = c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-1',false)">C) f is a polynomial</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-1',false)">D) f is increasing at x = c</button>

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<p><span class="label">Explanation:</span> Differentiability at a point always guarantees continuity at that same point.</p>

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<p class="prompt">2. If a function's graph has a sharp corner at x = c, what does this tell us about differentiability there?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-2',true)">A) f is not differentiable at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-2',false)">B) f is differentiable at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-2',false)">C) f is not continuous at c</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-2',false)">D) f has a horizontal tangent at c</button>

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<p><span class="label">Explanation:</span> At a corner, the slope approaching from the left disagrees with the slope approaching from the right, so no single derivative value exists there.</p>

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<p class="prompt">3. Is it possible for a function to be continuous at a point but not differentiable there?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-3',true)">A) Yes, continuity does not guarantee differentiability</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">B) No, continuity always implies differentiability</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">C) Only for polynomial functions</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-3',false)">D) Only at endpoints of a domain</button>

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<p><span class="label">Explanation:</span> f(x) = |x| is continuous at x = 0 but has a corner there, so it's not differentiable at that point. Differentiability implies continuity, but not the other way around.</p>

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<p class="prompt">4. A vertical tangent line at x = c indicates what about the derivative there?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-4',true)">A) f'(c) does not exist</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-4',false)">D) f(c) does not exist</button>

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<p><span class="label">Explanation:</span> A vertical tangent would require an infinite slope, so the derivative does not exist there, even though the function itself can still be continuous.</p>

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<p class="prompt">5. A cyclist's distance (miles) is recorded every 2 minutes: d(0) = 0, d(2) = 1.5, d(4) = 3.4, d(6) = 5.0. Estimate d'(2) using the symmetric difference quotient.</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p5-5',true)">A) 0.85 miles/min</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">B) 1.5 miles/min</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">C) 1.7 miles/min</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p5-5',false)">D) 0.425 miles/min</button>

</div>

<p><span class="label">Explanation:</span> d'(2) ≈ [d(4) − d(0)] / (2 × 2) = (3.4 − 0) / 4 = 0.85 miles per minute.</p>

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<h2 id="mvt">6. The Mean Value Theorem</h2>

<p>The <strong>Mean Value Theorem</strong> (MVT): if f is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is at least one c in (a, b) such that:</p>

<p><strong>Rolle's Theorem</strong> is a special case: if, in addition, f(a) = f(b), then there is a c in (a, b) where f'(c) = 0.</p>

<p><strong>Worked Example:</strong> You drive 240 miles in 4 hours. What does the Mean Value Theorem guarantee about your speed along the way?</p>

<p>Your average speed was 240 ÷ 4 = 60 mph. Since your instantaneous speed (the derivative of distance) existed continuously throughout the trip, the MVT guarantees that your speedometer read exactly 60 mph at least once.</p>

</div>

<p class="prompt">1. A runner covers 8 miles in 1 hour. By the Mean Value Theorem, what must be true about the runner's speed at some point during the run?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-1',true)">A) The runner's instantaneous speed equaled 8 mph at some point</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-1',false)">B) The runner's speed was always exactly 8 mph</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-1',false)">C) The runner never exceeded 8 mph</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-1',false)">D) The runner's average speed cannot be determined</button>

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<p><span class="label">Explanation:</span> The average speed (8 mph) must be matched by the instantaneous speed at least once, by the MVT.</p>

</div>

<p class="prompt">2. What are the two hypotheses required for the Mean Value Theorem to apply to f on [a, b]?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p6-2',true)">A) Continuous on [a, b] and differentiable on (a, b)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-2',false)">B) Differentiable on [a, b] only</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-2',false)">C) Continuous on (a, b) only</button>

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<p><span class="label">Explanation:</span> The MVT requires continuity on the closed interval and differentiability on the open interval, the endpoints themselves need not be differentiable.</p>

</div>

<p class="prompt">3. For f(x) = x² − 4x on [0, 6], find the value of c guaranteed by the Mean Value Theorem.</p>

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<p><span class="label">Explanation:</span> Average rate: [f(6) − f(0)] / 6 = (12 − 0)/6 = 2. Setting f'(x) = 2x − 4 equal to 2: 2x − 4 = 2, so x = 3.</p>

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<p class="prompt">4. Rolle's Theorem is a special case of the Mean Value Theorem that applies when…</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p6-4',false)">B) f is a polynomial</button>

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<p><span class="label">Explanation:</span> When f(a) = f(b), the average rate of change is 0, so Rolle's Theorem guarantees a point where the instantaneous rate (f') is also 0.</p>

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<h2 id="lhopital">7. L'Hospital's Rule and Recognizing a Limit as a Derivative</h2>

<p><strong>L'Hospital's Rule</strong>: if a limit of <span class="frac"><span class="num">f(x)</span><span class="den">g(x)</span></span> is indeterminate of the form <span class="frac"><span class="num">0</span><span class="den">0</span></span>, take the derivative of the numerator and denominator separately and try the limit again.</p>

Although several indeterminate forms exist in mathematics, only <span class="frac"><span class="num">0</span><span class="den">0</span></span> and <span class="frac"><span class="num">∞</span><span class="den">∞</span></span> are tested on the AP Calculus AB exam. Always confirm a limit is genuinely indeterminate before applying L'Hospital's Rule, using it elsewhere gives wrong answers.

</div>

<p>A limit can also sometimes be recognized directly as the definition of a derivative: lim<sub>h→0</sub> [f(c + h) − f(c)]/h is exactly f'(c) for some function f and value c.</p>

<p><strong>Worked Example:</strong> Find lim<sub>x→2</sub> <span class="frac"><span class="num">x² − 4</span><span class="den">x − 2</span></span> using L'Hospital's Rule.</p>

<p>This is of the form 0/0. Differentiating numerator and denominator separately:</p>

</div>

<p class="prompt">1. Find lim<sub>x→0</sub> <span class="frac"><span class="num">sin(4x)</span><span class="den">x</span></span> using L'Hospital's Rule.</p>

</div>

<p><span class="label">Explanation:</span> This is 0/0. Differentiating: <span class="frac"><span class="num">4cos(4x)</span><span class="den">1</span></span>, which at x = 0 gives 4(1) = 4.</p>

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<p><span class="label">Explanation:</span> This is ∞/∞. Differentiating: <span class="frac"><span class="num">1/x</span><span class="den">2x</span></span> = <span class="frac"><span class="num">1</span><span class="den">2x²</span></span>, which approaches 0 as x → ∞.</p>

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<p><span class="label">Explanation:</span> This is 0/0. Differentiating: <span class="frac"><span class="num">3e³<sup>x</sup></span><span class="den">1</span></span>, which at x = 0 gives 3(1) = 3.</p>

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<p class="prompt">4. Which indeterminate forms are eligible for L'Hospital's Rule as tested on the AP Calculus AB exam?</p>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-4',false)">B) All indeterminate forms</button>

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<p><span class="label">Explanation:</span> While L'Hospital's Rule can theoretically be extended to other indeterminate forms with algebraic manipulation, the AP exam only tests it directly on 0/0 and ∞/∞.</p>

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<p class="prompt">5. The limit lim<sub>h→0</sub> <span class="frac"><span class="num">(5 + h)³ − 5³</span><span class="den">h</span></span> is the derivative of which function, evaluated at which point, and what is its value?</p>

<button class="option-btn" data-correct="true" onclick="checkAnswer(this,'p7-5',true)">A) f(x) = x³, evaluated at x = 5 (value: 75)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">B) f(x) = x³, evaluated at x = 3 (value: 27)</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">C) f(x) = 5x³, evaluated at x = 0</button>

<button class="option-btn" data-correct="false" onclick="checkAnswer(this,'p7-5',false)">D) f(x) = x⁵, evaluated at x = 3</button>

</div>

<p><span class="label">Explanation:</span> This matches the definition f'(c) = lim<sub>h→0</sub> [f(c+h) − f(c)]/h with f(x) = x³ and c = 5. Since f'(x) = 3x², f'(5) = 3(25) = 75.</p>

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<h2 id="mistakes">Common Mistakes to Avoid</h2>

<li><strong>Forgetting the Chain Rule factor when differentiating a composition.</strong> Whenever the "inside" of a function isn't simply x, an extra factor (the inside function's own derivative) must be multiplied in, this is one of the single most common AP scoring deductions.</li>

<li><strong>Mixing up the order of terms in the Quotient Rule.</strong> It's "bottom times derivative of top, minus top times derivative of bottom," reversing this order flips the sign of the answer.</li>

<li><strong>Forgetting to differentiate y-terms with an extra dy/dx factor during implicit differentiation.</strong> Every time you differentiate a term containing y, the Chain Rule requires multiplying by dy/dx, since y is itself a function of x.</li>

<li><strong>Applying L'Hospital's Rule to a limit that isn't actually indeterminate.</strong> Confirm the limit truly evaluates to 0/0 or ∞/∞ by direct substitution first, using the rule on a limit that's already a real number gives a wrong answer.</li>

<li><strong>Assuming continuity is enough to guarantee differentiability.</strong> A function can be perfectly continuous at a point and still fail to have a derivative there, due to a corner, cusp, or vertical tangent.</li>

</ul>

<h2 id="faq">Frequently Asked Questions</h2>

<h3>How can I tell whether to use the Product Rule or the Chain Rule?</h3>

<p>The Product Rule applies when two separate functions are being multiplied together, like x² · sin x. The Chain Rule applies when one function is nested inside another, like sin(x²). Some problems, like (x²+1)sin(2x), genuinely need both rules working together.</p>

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<h3>Why does implicit differentiation need the extra dy/dx factor?</h3>

<p>Since y is being treated as an unknown function of x rather than an independent variable, every time the Chain Rule is applied to a y-term, it must include the derivative of y with respect to x, exactly as it would for any other composite function.</p>

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<h3>Is Rolle's Theorem tested by name on the AP exam?</h3>

<p>No, but the underlying idea, that a continuous, differentiable function returning to the same value must have a horizontal tangent somewhere in between, is still worth understanding as a special case of the more general Mean Value Theorem.</p>

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<h3>Where can I practice more problems like these?</h3>

<p>The <a href="https://theschoolofmathematics.com/quiz/differentiation-quiz-1">Differentiation quizzes</a> in the AP Calculus AB Question Bank include additional original problems on this topic, along with quizzes covering every other topic tested throughout the course.</p>

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<a class="cta-btn cta-primary" href="https://theschoolofmathematics.com/quiz/differentiation-quiz-1">Practice Differentiation Free</a>

<a class="cta-btn cta-secondary" href="https://theschoolofmathematics.com/quiz/course/AP-Calculus-AB-QBank">Explore the Full AP Calculus AB Qbank</a>

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