A rigorous, publication-grade worksheet for students who want more than mechanical practice — designed to build deep structural understanding of rational function differentiation.
The history of calculus is inseparable from the study of ratios. Long before Leibniz formalized the notation \(\frac{dy}{dx}\), natural philosophers spoke of rates as comparisons — how fast does distance accumulate relative to time, how rapidly does pressure change relative to volume. Division was always at the heart of the question. What makes the Quotient Rule remarkable is not that it exists, but that it exists in exactly the asymmetric, order-sensitive form it does. Students who memorize the formula without understanding why it is structured the way it is will make sign errors for years. Students who understand the derivation will almost never make that mistake again.
The Quotient Rule emerges naturally from the Product Rule. If \(h(x) = f(x)/g(x)\), we can write \(f(x) = h(x) \cdot g(x)\) and differentiate both sides. Applying the Product Rule to the right side and solving for \(h'(x)\) yields exactly the formula you will use throughout this worksheet. This is not a historical footnote — it is the single most useful way to remember and reconstruct the formula under pressure, when memory fails at the worst moment.
Intuitively, consider what happens when you differentiate a ratio. The numerator is changing, pulling the output one direction. The denominator is also changing, scaling the whole expression by something that is itself shifting. The two effects interact, and neither is simply additive. If the denominator grows while the numerator stays constant, the ratio shrinks — even though no subtraction occurred in the inputs. This coupling between numerator and denominator dynamics is exactly what the formula captures, with the squared denominator term encoding how strongly the denominator's movement "stretches" the ratio at each point.
In practice, the Quotient Rule is ubiquitous in applied mathematics. Rational functions — expressions of the form \(p(x)/q(x)\) — appear as models of drug concentration in the bloodstream (where the body metabolizes a substance at a rate proportional to its current level), as efficiency curves in mechanical engineering (torque-to-RPM ratios), and as price elasticity functions in economics (the ratio of percentage change in demand to percentage change in price). Every time you encounter a ratio whose two components both depend on the same underlying variable, the Quotient Rule is the appropriate differentiation tool.
This worksheet is designed for students who have already encountered the Power Rule and Product Rule, and who are ready to work with rational expressions, trigonometric ratios, and functions where multiple differentiation rules interact. It is equally appropriate for a motivated precalculus student beginning to study limits, a student in AP Calculus AB or BC preparing for the exam, or a first-year undergraduate refining their differentiation technique ahead of multivariable calculus. The problems here are arranged not merely by difficulty but by structural type — so that you build genuine pattern recognition, not just computational speed.
Mastery of the Quotient Rule looks like this: you see a ratio, you correctly identify numerator and denominator without hesitation, you apply the formula maintaining the correct order of subtraction, and you simplify the result efficiently. More than that, mastery means knowing when not to use the Quotient Rule — recognizing that \(\frac{x^3 + 2x}{x}\) is better handled by splitting the fraction first, or that \(\frac{1}{x^2 + 1}\) is sometimes cleaner via the Chain Rule on \((x^2+1)^{-1}\). The goal of this worksheet is to develop exactly that kind of contextual judgment, rather than a reflexive formula-application that breaks down the moment a problem presents itself in an unfamiliar form.
If \(f\) and \(g\) are differentiable at \(x\) and \(g(x) \neq 0\), then:
\[\frac{d}{dx}\!\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{[g(x)]^2}\]Order matters: numerator of the formula is always top' × bottom − top × bottom', never the reverse.
The cleanest derivation uses the Product Rule. Let \(h(x) = f(x)/g(x)\), so \(f(x) = h(x) \cdot g(x)\). Differentiating both sides:
\[f'(x) = h'(x)\,g(x) + h(x)\,g'(x)\]Substituting \(h(x) = f(x)/g(x)\) and solving for \(h'(x)\):
\[h'(x) = \frac{f'(x) - h(x)\,g'(x)}{g(x)} = \frac{f'(x) - \tfrac{f(x)}{g(x)}\,g'(x)}{g(x)} = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{[g(x)]^2}\]This derivation reveals something important: the formula is not arbitrary. The squared denominator arises because we divided through by \(g(x)\) once — and \(g(x)\) was already sitting as a factor in the denominator. Whenever you wonder why it's \(g(x)^2\) and not just \(g(x)\), recall this derivation.
| Structure | Which Rule | Reason |
|---|---|---|
| \(\dfrac{x^2+1}{3}\) | Power / Constant | Constant denominator — just divide coefficients |
| \(\dfrac{x^3}{x+2}\) | Quotient Rule | Both numerator and denominator depend on \(x\) |
| \(\dfrac{\sin x}{x^2}\) | Quotient Rule | Trig numerator, polynomial denominator |
| \(\dfrac{1}{u(x)}\) | Chain Rule on \(u^{-1}\) | Often simpler than Quotient Rule |
| \(\dfrac{e^x \sin x}{x+1}\) | Quotient + Product | Numerator requires Product Rule first |
| \(\dfrac{f(g(x))}{h(x)}\) | Quotient + Chain | Numerator is composite — Chain Rule applies there |
Not every fraction demands the Quotient Rule. Develop the habit of asking: can I simplify first? A function like \(\frac{x^4 - 3x^2 + x}{x}\) simplifies term-by-term to \(x^3 - 3x + 1\), after which ordinary power rule differentiation takes three seconds. Forcing the Quotient Rule on this expression wastes time and introduces unnecessary opportunities for algebraic error. Similarly, \(\frac{5}{x^3}\) is more naturally written as \(5x^{-3}\), differentiated instantly by the Power Rule. Reserve the Quotient Rule for situations where such simplification is genuinely impossible.
Find \(\dfrac{d}{dx}\!\left[\dfrac{x^2 + 3}{x - 1}\right]\)
Step 1 — Identify f and g
\(f(x) = x^2 + 3\), so \(f'(x) = 2x\).
\(g(x) = x - 1\), so \(g'(x) = 1\).
Step 2 — Apply the formula
\(\dfrac{f'g - fg'}{g^2} = \dfrac{2x(x-1) - (x^2+3)(1)}{(x-1)^2}\)
Step 3 — Expand numerator
\(= \dfrac{2x^2 - 2x - x^2 - 3}{(x-1)^2} = \dfrac{x^2 - 2x - 3}{(x-1)^2}\)
Step 4 — Factor if possible
\(x^2 - 2x - 3 = (x-3)(x+1)\), so the final answer factors neatly.
Common Mistake
Squaring only the derivative of the denominator: writing \(1^2 = 1\) in the denominator rather than \((x-1)^2\).
Find \(\dfrac{d}{dx}\!\left[\dfrac{\sin x}{x^2 + 1}\right]\)
Step 1 — Identify f and g
\(f(x) = \sin x \Rightarrow f'(x) = \cos x\).
\(g(x) = x^2 + 1 \Rightarrow g'(x) = 2x\).
Step 2 — Apply the formula
\(\dfrac{\cos x (x^2+1) - \sin x \cdot 2x}{(x^2+1)^2}\)
Step 3 — Simplify
The numerator does not factor further in elementary form; leave as is.
Common Mistake
Differentiating \(\sin x\) as \(-\cos x\) (confusing with \(\cos x \to -\sin x\) direction).
Find \(\dfrac{d}{dx}\!\left[\dfrac{e^{3x}}{x^2 + 5}\right]\)
Step 1 — Identify f and g; note chain rule needed on f
\(f(x) = e^{3x}\). By the Chain Rule: \(f'(x) = 3e^{3x}\).
\(g(x) = x^2 + 5 \Rightarrow g'(x) = 2x\).
Step 2 — Apply Quotient Rule
\(\dfrac{3e^{3x}(x^2+5) - e^{3x}\cdot 2x}{(x^2+5)^2}\)
Step 3 — Factor the numerator
\(= \dfrac{e^{3x}[3(x^2+5) - 2x]}{(x^2+5)^2} = \dfrac{e^{3x}(3x^2 - 2x + 15)}{(x^2+5)^2}\)
Why factor?
Factoring \(e^{3x}\) from the numerator is a simplification graders appreciate and it makes the expression easier to use in subsequent calculations (e.g., finding critical points).
Find \(\dfrac{d}{dx}\!\left[\dfrac{x^2 \ln x}{x + 1}\right]\)
Step 1 — Numerator requires the Product Rule
\(f(x) = x^2 \ln x\). Apply Product Rule:
\(f'(x) = 2x \ln x + x^2 \cdot \dfrac{1}{x} = 2x\ln x + x\).
Step 2 — Denominator
\(g(x) = x+1 \Rightarrow g'(x) = 1\).
Step 3 — Quotient Rule
\(\dfrac{(2x\ln x + x)(x+1) - x^2\ln x \cdot 1}{(x+1)^2}\)
Step 4 — Expand the numerator carefully
\(= \dfrac{2x(x+1)\ln x + x(x+1) - x^2\ln x}{(x+1)^2}\)
\(= \dfrac{\ln x[2x^2 + 2x - x^2] + x^2 + x}{(x+1)^2} = \dfrac{(x^2+2x)\ln x + x^2+x}{(x+1)^2}\)
Simplify further
\(= \dfrac{x(x+2)\ln x + x(x+1)}{(x+1)^2} = \dfrac{x[(x+2)\ln x + (x+1)]}{(x+1)^2}\)
Common Mistake Warning
Forgetting to apply the Product Rule to the numerator before inserting it into the Quotient Rule formula. This is the single most common multi-rule error at the AP level.
Find \(\dfrac{d}{dx}\!\left[\dfrac{\cos x}{x^3}\right]\) using two methods.
Method A — Quotient Rule
\(f = \cos x,\ f' = -\sin x;\ g = x^3,\ g' = 3x^2\).
\(\dfrac{-\sin x \cdot x^3 - \cos x \cdot 3x^2}{x^6} = \dfrac{-x^2(x\sin x + 3\cos x)}{x^6} = \dfrac{-(x\sin x + 3\cos x)}{x^4}\)
Method B — Rewrite as Product
\(\dfrac{\cos x}{x^3} = \cos x \cdot x^{-3}\). Apply Product Rule:
\(-\sin x \cdot x^{-3} + \cos x \cdot (-3)x^{-4} = \dfrac{-\sin x}{x^3} - \dfrac{3\cos x}{x^4}\)
Combine over \(x^4\): \(\dfrac{-x\sin x - 3\cos x}{x^4}\). Same answer.
Which method is faster?
Method B is often faster for simple denominators that are pure power functions. Choose based on structure, not habit.
Find \(\dfrac{d}{dx}\!\left[\dfrac{4x^3 - 8x}{4x}\right]\)
Trap: Applying Quotient Rule immediately
A mechanical student writes \(f = 4x^3 - 8x,\ f' = 12x^2 - 8;\ g = 4x,\ g' = 4\) and begins a lengthy calculation. This is unnecessary.
Smart approach: Simplify first
\(\dfrac{4x^3 - 8x}{4x} = \dfrac{4x(x^2 - 2)}{4x} = x^2 - 2\) (for \(x \neq 0\))
Then: \(\dfrac{d}{dx}[x^2 - 2] = 2x\). Done.
Lesson
Always ask whether a fraction simplifies before reaching for a differentiation rule. This habit saves time on timed exams and reduces error.
Find \(\dfrac{d}{dx}\!\left[\dfrac{x^2}{\sin x + \cos x}\right]\)
Step 1 — Identify parts
\(f = x^2,\ f' = 2x\); \(g = \sin x + \cos x,\ g' = \cos x - \sin x\).
Step 2 — Apply formula
\(\dfrac{2x(\sin x + \cos x) - x^2(\cos x - \sin x)}{(\sin x + \cos x)^2}\)
Step 3 — Expand and group
\(= \dfrac{x[2(\sin x + \cos x) - x(\cos x - \sin x)]}{(\sin x + \cos x)^2}\)
\(= \dfrac{x[(2+x)\sin x + (2-x)\cos x]}{(\sin x + \cos x)^2}\)
Note on domain
This expression is undefined where \(\sin x + \cos x = 0\), i.e., where \(\tan x = -1\), meaning \(x = -\pi/4 + n\pi\) for integer \(n\).
Find \(\dfrac{d}{dx}\!\left[\dfrac{\sqrt{x^2+1}}{x\,e^x}\right]\)
Step 1 — Numerator via Chain Rule
\(f = (x^2+1)^{1/2} \Rightarrow f' = \dfrac{x}{\sqrt{x^2+1}}\)
Step 2 — Denominator via Product Rule
\(g = xe^x \Rightarrow g' = e^x + xe^x = e^x(1+x)\)
Step 3 — Quotient Rule
\(\dfrac{\dfrac{x}{\sqrt{x^2+1}} \cdot xe^x - \sqrt{x^2+1}\cdot e^x(1+x)}{x^2 e^{2x}}\)
Step 4 — Factor \(e^x\) from numerator
\(= \dfrac{e^x\!\left[\dfrac{x^2}{\sqrt{x^2+1}} - (1+x)\sqrt{x^2+1}\right]}{x^2 e^{2x}}\)
Step 5 — Combine fractions in numerator over \(\sqrt{x^2+1}\)
Numerator of bracket: \(\dfrac{x^2 - (1+x)(x^2+1)}{\sqrt{x^2+1}} = \dfrac{x^2 - x^3 - x - x^2 - 1}{\sqrt{x^2+1}} = \dfrac{-x^3 - x - 1}{\sqrt{x^2+1}}\)
Step 6 — Simplify overall
\(= \dfrac{-e^x(x^3 + x + 1)}{x^2 e^{2x}\sqrt{x^2+1}} = \dfrac{-(x^3+x+1)}{x^2 e^x \sqrt{x^2+1}}\)
Reflection
Multi-rule problems rarely simplify completely. Work step-by-step; don't try to hold more than one level of complexity in your head at once.
Work through each section in order. Each section builds on the structural skills developed in the previous one. Resist the temptation to skip to later sections — the early problems train the mechanical habits that protect against errors in advanced problems.
Apply the Quotient Rule directly to polynomial and simple algebraic ratios. All functions here have straightforward derivatives for numerator and denominator. Focus on maintaining correct subtraction order and squaring the full denominator.
For each problem, first decide which differentiation strategy is most efficient (Quotient Rule, algebraic simplification, or rewriting as a negative power). State your strategy, then execute.
These problems require combining the Quotient Rule with at least one other rule. Identify all required rules before beginning each problem.
These problems ground the Quotient Rule in real contexts. For each, find the derivative, interpret its units, and explain what the value tells you about the situation at the given point.
These problems reflect the format and difficulty of AP Calculus AB/BC and standard university Calculus I final exams. Time yourself: aim for 3–5 minutes per problem. Multiple choice options are provided for some; show work regardless.
These problems are designed to stretch beyond standard exam expectations. They reward structural insight, careful algebra, and comfort with non-routine forms. Each has at least one insight that makes the problem tractable — find it before computing blindly.
For each function below, do not differentiate immediately. First, write down which rule(s) apply and justify your reasoning in one sentence. Then compute the derivative.
These problems do not require computation. They test understanding of the Quotient Rule at a deeper level — the kind of understanding that allows you to reconstruct facts under pressure.
The most common Quotient Rule error, by a wide margin: writing \(fg' - f'g\) instead of \(f'g - fg'\) in the numerator. The Quotient Rule is not symmetric. The Product Rule is (fg)' = f'g + fg' — both terms are added. But the Quotient Rule involves subtraction, and that subtraction has a specific order. If you ever feel uncertain, rederive from the Product Rule in thirty seconds. Do not try to remember it through brute memorization alone.
The denominator of the Quotient Rule formula is \([g(x)]^2\) — the entire denominator squared. Not \([g'(x)]^2\). A surprising number of students, under exam pressure, write the derivative of the denominator squared rather than the original denominator squared. Awareness of this specific error cuts its frequency dramatically.
When the numerator or denominator is itself a composite function — \(e^{3x}\), \(\sin(x^2)\), \((x^2+1)^4\) — you must apply the Chain Rule when computing \(f'(x)\) or \(g'(x)\). Forgetting this step produces an answer that appears plausible but is subtly wrong. Always inspect the numerator and denominator for composition before writing any derivatives.
AP free-response problems frequently give a table of values for \(f\), \(f'\), \(g\), \(g'\) at specific points and ask for \(h'(a)\) where \(h = f/g\). The error here is not algebraic — it's organizational. Carefully read which values correspond to which functions and which \(x\)-values. A single misread from the table produces a completely wrong answer that earns no credit even if the formula was applied correctly.
On university exams, an unsimplified answer often receives partial credit. On AP free-response, it may receive full credit — but on multiple-choice, an unsimplified form will not match any listed option. Develop the discipline of always factoring a common term from the numerator after applying the Quotient Rule. This also helps you find critical points, since setting the numerator equal to zero (where the denominator is nonzero) identifies where the derivative is zero.
At the AP level, free-response graders apply a "method-error" distinction: if you set up the Quotient Rule correctly but make an arithmetic error in expansion, you lose at most one point for the error itself but keep all method points. However, if you apply the wrong formula — using the Product Rule where the Quotient Rule was needed, or reversing the subtraction sign — you lose all method points for that sub-part. This means: formula accuracy matters more than arithmetic speed. Write the formula explicitly, plug in carefully, then expand.
At the university level, the calculus of partial credit is similar but graders tend to be less lenient about formula errors. A wrong sign in the setup receives zero for the entire problem at many institutions. Write slowly and check the subtraction order before expanding anything.
When you see a quotient on a timed exam, spend ten seconds checking: (1) Can I simplify algebraically? (2) Is the denominator a pure power of \(x\), making negative-exponent rewriting cleaner? (3) If not, apply Quotient Rule. This ten-second scan prevents you from spending two minutes applying the Quotient Rule to something like \(\frac{x^4 - x^2}{x^2} = x^2 - 1\).
Answers are provided with key steps. For brevity, intermediate algebra is condensed but logical flow is preserved. For sections D and E with reasoning, key interpretations are noted.
\(f = x^3,\ f' = 3x^2;\ g = x+4,\ g' = 1\)
\(= \dfrac{3x^2(x+4) - x^3(1)}{(x+4)^2} = \dfrac{3x^3 + 12x^2 - x^3}{(x+4)^2} = \dfrac{2x^3 + 12x^2}{(x+4)^2}\)
\(f = 2x+5,\ f' = 2;\ g = x^2-1,\ g' = 2x\)
\(= \dfrac{2(x^2-1) - (2x+5)(2x)}{(x^2-1)^2} = \dfrac{2x^2 - 2 - 4x^2 - 10x}{(x^2-1)^2} = \dfrac{-2x^2 - 10x - 2}{(x^2-1)^2}\)
Simplify first: \(\dfrac{x^2-4}{x+2} = \dfrac{(x-2)(x+2)}{x+2} = x-2\) for \(x \neq -2\).
Derivative: \(1\).
\(f = 3x^4,\ f' = 12x^3;\ g = 2x^3+1,\ g' = 6x^2\)
\(= \dfrac{12x^3(2x^3+1) - 3x^4 \cdot 6x^2}{(2x^3+1)^2} = \dfrac{24x^6 + 12x^3 - 18x^6}{(2x^3+1)^2} = \dfrac{6x^6 + 12x^3}{(2x^3+1)^2}\)
\(f = 1-x^2,\ f' = -2x;\ g = 1+x^2,\ g' = 2x\)
\(= \dfrac{-2x(1+x^2) - (1-x^2)(2x)}{(1+x^2)^2} = \dfrac{-2x - 2x^3 - 2x + 2x^3}{(1+x^2)^2} = \dfrac{-4x}{(1+x^2)^2}\)
\(f = x^3+2x,\ f' = 3x^2+2;\ g = x^2+x+1,\ g' = 2x+1\)
Numerator: \((3x^2+2)(x^2+x+1) - (x^3+2x)(2x+1)\)
\(= 3x^4+3x^3+3x^2+2x^2+2x+2 - (2x^4+x^3+4x^2+2x)\)
\(= x^4 + 2x^3 + x^2 + 2\)
\(f = 5x-3,\ f' = 5;\ g = 4x+7,\ g' = 4\)
\(= \dfrac{5(4x+7) - (5x-3)(4)}{(4x+7)^2} = \dfrac{20x+35-20x+12}{(4x+7)^2} = \dfrac{47}{(4x+7)^2}\)
Simplify: \(\dfrac{x^2+x-6}{x-2} = \dfrac{(x+3)(x-2)}{x-2} = x+3\) for \(x \neq 2\). Derivative: \(1\).
\(f = x^5,\ f' = 5x^4;\ g = x^2+3,\ g' = 2x\)
\(= \dfrac{5x^4(x^2+3) - x^5 \cdot 2x}{(x^2+3)^2} = \dfrac{5x^6+15x^4 - 2x^6}{(x^2+3)^2} = \dfrac{3x^6+15x^4}{(x^2+3)^2}\)
\(f = x^{1/2},\ f' = \tfrac{1}{2}x^{-1/2};\ g = x+1,\ g' = 1\)
\(= \dfrac{\frac{1}{2\sqrt{x}}(x+1) - \sqrt{x}}{(x+1)^2} = \dfrac{\frac{x+1-2x}{2\sqrt{x}}}{(x+1)^2} = \dfrac{1-x}{2\sqrt{x}(x+1)^2}\)
Simplify: \(\dfrac{6x^3-3x}{3x} = \dfrac{3x(2x^2-1)}{3x} = 2x^2-1\). Derivative: \(4x\).
\(f = \sin x,\ f' = \cos x;\ g = x,\ g' = 1\)
\(= \dfrac{x\cos x - \sin x}{x^2}\)
\(= \dfrac{e^x \cdot x^2 - e^x \cdot 2x}{x^4} = \dfrac{e^x(x-2)}{x^3}\)
\(f' = 1/x;\ g' = 1\). \(= \dfrac{\frac{1}{x}\cdot x - \ln x \cdot 1}{x^2} = \dfrac{1 - \ln x}{x^2}\)
\(f' = \sec^2 x;\ g' = 1\). \(= \dfrac{(x+1)\sec^2 x - \tan x}{(x+1)^2}\)
Numerator product rule: \(f = x^2 e^x,\ f' = 2xe^x + x^2 e^x = xe^x(2+x)\).
\(g = x+3,\ g' = 1\).
Numerator of QR: \(xe^x(x+2)(x+3) - x^2 e^x = e^x[x(x+2)(x+3) - x^2]\)
\(= e^x[x(x^2+5x+6) - x^2] = e^x[x^3 + 5x^2 + 6x - x^2] = e^x(x^3+4x^2+6x)\)
Chain Rule on numerator: \(f = \sin^2 x,\ f' = 2\sin x \cos x = \sin(2x)\).
\(g = x^2+1,\ g' = 2x\).
\(= \dfrac{\sin(2x)(x^2+1) - 2x\sin^2 x}{(x^2+1)^2}\)
\(C(t) = \dfrac{80t}{t^2+4}\). \(f' = 80,\ g' = 2t\).
\(C'(t) = \dfrac{80(t^2+4) - 80t \cdot 2t}{(t^2+4)^2} = \dfrac{80(4-t^2)}{(t^2+4)^2}\)
At \(t=2\): \(C'(2) = \dfrac{80(4-4)}{64} = 0\). The concentration is at its maximum at \(t=2\) hours — the drug is neither accumulating nor clearing at this instant.
\(s(t) = \dfrac{t^2-1}{t^2+1}\). \(f' = 2t,\ g' = 2t\).
\(v(t) = \dfrac{2t(t^2+1) - (t^2-1)(2t)}{(t^2+1)^2} = \dfrac{2t[(t^2+1)-(t^2-1)]}{(t^2+1)^2} = \dfrac{4t}{(t^2+1)^2}\)
For \(t>0\): \(4t>0\) and \((t^2+1)^2 > 0\), so \(v(t) > 0\). Particle always moves right.
\(v(1) = \dfrac{4}{4} = 1\) m/s.
\(h'(x) = \dfrac{f'g - fg'}{g^2}\)
\(h'(2) = \dfrac{(-1)(1) - (3)(4)}{1^2} = \dfrac{-1-12}{1} = -13\)
\(y = \dfrac{2x}{x^2+1}\). At \(x=1\): \(y = 1\).
\(y' = \dfrac{2(x^2+1) - 2x\cdot 2x}{(x^2+1)^2} = \dfrac{2-2x^2}{(x^2+1)^2}\)
\(y'(1) = \dfrac{2-2}{4} = 0\). Tangent line is horizontal.
\(f = x^3,\ f' = 3x^2;\ g = e^x,\ g' = e^x\).
\(= \dfrac{3x^2 e^x - x^3 e^x}{e^{2x}} = \dfrac{e^x(3x^2-x^3)}{e^{2x}} = \dfrac{x^2(3-x)}{e^x}\)
\(f = \sec x,\ f' = \sec x \tan x;\ g = 1+\tan x,\ g' = \sec^2 x\)
Numerator: \(\sec x \tan x(1+\tan x) - \sec x \cdot \sec^2 x = \sec x[\tan x + \tan^2 x - \sec^2 x]\)
Since \(\tan^2 x - \sec^2 x = -1\):
\(= \sec x(\tan x - 1)\)
First: \(y' = \dfrac{(x^2+1) - x(2x)}{(x^2+1)^2} = \dfrac{1-x^2}{(x^2+1)^2}\)
Second: Apply QR again. \(f = 1-x^2,\ f' = -2x;\ g = (x^2+1)^2,\ g' = 4x(x^2+1)\)
\(y'' = \dfrac{-2x(x^2+1)^2 - (1-x^2)\cdot 4x(x^2+1)}{(x^2+1)^4}\)
\(= \dfrac{-2x(x^2+1) - 4x(1-x^2)}{(x^2+1)^3} = \dfrac{-2x^3-2x-4x+4x^3}{(x^2+1)^3} = \dfrac{2x^3-6x}{(x^2+1)^3} = \dfrac{2x(x^2-3)}{(x^2+1)^3}\)
Inflection: \(y'' = 0\) when \(x = 0\) or \(x = \pm\sqrt{3}\).
The student wrote \(\dfrac{2x}{1} = 2x\). Errors:
(1) Treated denominator as constant 1 — but \(g = x-1\) so \(g(x)\) is not 1; the student conflated \(g'(x)=1\) with \(g(x)=1\).
(2) Did not square the denominator — even if the formula had been applied, the denominator must be \((x-1)^2\), not \(1\).
(3) Omitted the \(-f(x)g'(x)\) term entirely.
Correct answer: \(\dfrac{(x-3)(x+1)}{(x-1)^2}\) (see Example 1).
The student wrote \(\dfrac{\cos x}{2x}\). This is the ratio of derivatives — it applies no differentiation rule at all. They differentiated numerator and denominator independently and divided, which is not a valid rule. The correct approach uses the Quotient Rule: \(\dfrac{x^2\cos x - \sin x \cdot 2x}{x^4} = \dfrac{x\cos x - 2\sin x}{x^3}\).
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