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♾️AP Calculus AB/BC Unit 2 Review

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2.9 The Quotient Rule

♾️AP Calculus AB/BC
Unit 2 Review

2.9 The Quotient Rule

Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025
♾️AP Calculus AB/BC
Unit & Topic Study Guides
Unit 2 Overview: Differentiation
2.1 Defining Average and Instantaneous Rates of Change at a Point
2.2 Defining the Derivative of a Function and Using Derivative Notation
2.3 Estimating Derivatives of a Function at a Point
2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist
2.5 Applying the Power Rule
2.6 Derivative Rules: Constant, Sum, Difference, and Constant Multiple
2.7 Derivatives of cos x, sinx, e^x, and ln x
2.8 The Product Rule
2.9 The Quotient Rule
2.10 Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
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2.9 The Quotient Rule

As the complexity of the functions you encounter increases, applying the basic Power Rule correctly becomes more difficult. Like the Product Rule explored in the previous lesson, the Quotient Rule simplifies the process of differentiating the quotient of functions, allowing for the differentiation of larger and more complex functions. 📈

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⛳ Quotient Rule Definition

The Quotient Rule is as stated below:

ddx[f(x)g(x)]=g(x)ddxf(x)−f(x)ddxg(x)(g(x))2\frac{d}{dx} \Bigg[\frac{f(x)}{g(x)}\Bigg] = \frac{g(x)\frac{d}{dx}f(x)-f(x)\frac{d}{dx}g(x)}{(g(x))^2}

Similar to the Product Rule, the Quotient Rule uses the original functions and their derivatives to compute the final derivative. Rather than adding the products of the functions and their derivatives, they are subtracted and divided by the square of the denominator function (in this case, g(x)g(x)). If this sounds and looks a little too complicated, you can substitute variables to represent the functions as shown below.

Given that u=f(x)\textcolor{red}{u = f(x)} and v=g(x)\textcolor{green}{v = g(x)}, then the Quotient Rule can also be written as:

ddx[f(x)g(x)]=ddx(uv)=vu′−uv′v2\frac{d}{dx} \Bigg[\frac{\textcolor{red}{f(x)}}{\textcolor{green}{g(x)}}\Bigg] = \frac{d}{dx} (\frac{\textcolor{red}{u}}{\textcolor{green}{v}}) = \frac{\textcolor{green}{v}\textcolor{purple}{u'}-\textcolor{red}{u}\textcolor{blue}{v'}}{\textcolor{green}{v^2}}

Note: The Quotient Rule can only be used when the equation is in the form of f(x)g(x)\frac{f(x)}{g(x)} AND both functions are differentiable. ✅

✏️ Quotient Rule Walkthrough

Now that we know what the Quotient Rule is and how to use it, let's apply it to an example! 📚

🔢 Suppose you are given the rational function shown below:

y=x2+x−2x3+6y=\frac{x^2+x-2}{x^3+6}

First, notice how it is already in the form of f(x)g(x)\frac{f(x)}{g(x)}, so we can apply the Quotient Rule to calculate its derivative. If we say that f(x)=x2+x−2f(x) = x^2+x-2 and g(x)=x3+6g(x) = x^3 + 6, then:

dydx[x2+x−2x3+6]=(x3+6)∗ddx(x2+x−2)−(x2+x−2)∗ddx(x3+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)*\frac{d}{dx}(x^2+x-2)-(x^2+x-2)*\frac{d}{dx}(x^3+6)}{(x^3+6)^2}

Using the formula given by the Quotient Rule, the derivative of this rational function is rewritten, and can be simplified as follows: ✍️

dydx[x2+x−2x3+6]=(x3+6)(2x+1)−(3x2)(x2+x−2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(x^3+6)(2x+1)-(3x^2)(x^2+x-2)}{(x^3+6)^2}

Here, we rewrote the rational function with the derivatives of the numerator and denominator. After this, you can use algebra to combine and further simplify the derivative:

dydx[x2+x−2x3+6]=(2x4+x3+12x+6)−(3x4+3x3−6x2)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] = \frac{(2x^4+x^3+12x+6)-(3x^4+3x^3-6x^2)}{(x^3+6)^2}

By multiplying binomial and combining like terms, our final answer is:

dydx[x2+x−2x3+6]=(−x4−2x3+6x2+12x+6)(x3+6)2\frac{dy}{dx} \Bigg[\frac{x^2+x-2}{x^3+6}\Bigg] =\frac{(-x^4-2x^3+6x^2+12x+6)}{(x^3+6)^2}

From here, the derivative can be simplified further, but this is only sometimes necessary. Do try to simplify the derivative if possible. 💡

This rule can also be applied to other types of functions besides algebraic functions, such as trigonometric functions (sin⁡(x),cos⁡(x),tan⁡(x)\sin(x), \cos(x), \tan(x) , etc) and exponential functions (ex)(e^x). Try the next two examples of applying the Quotient Rule to differentiate these types of functions! 🏁

🧮 Quotient Rule: Practice Problems

Give the following problems a try yourself before you see the way we walk through them!

Quotient Rule: Example 1

Suppose you are given the rational function below:

y=ex1+x2y = \frac{e^x}{1+x^2}

In applying the Quotient Rule, we see that f(x)=exf(x) = e^x and g(x)=1+x2g(x) = 1+x^2. Recall that the derivative of exe^x is just exe^x.

dydx[ex1+x2]=(1+x2)∗ddx(ex)−(ex)∗ddx(1+x2)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*\frac{d}{dx}(e^x)-(e^x)*\frac{d}{dx}(1+x^2)}{(1+x^2)^2}

Therefore,

dydx[ex1+x2]=(1+x2)∗(ex)−(ex)∗(2x)(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{(1+x^2)*(e^x)-(e^x)*(2x)}{(1+x^2)^2}

This can either be left as is, but can also be simplified for better clarity:

dydx[ex1+x2]=ex(1−2x+x2)(1+x2)2=ex(1−x)2(1+x2)2\frac{dy}{dx} \Bigg[\frac{e^x}{1+x^2}\Bigg] = \frac{e^x(1-2x+x^2)}{(1+x^2)^2} = \frac{e^x(1-x)^2}{(1+x^2)^2}

This makes the final answer to be ex(1−x)2(1+x2)2\frac{e^x(1-x)^2}{(1+x^2)^2}! Now let’s try using the Quotient Rule with trigonometric functions. ⬇️

Quotient Rule: Example 2

Suppose you are given the following rational function below:

y=sin(x)1+cos(x)y = \frac{sin(x)}{1+cos(x)}

Again, as we apply the Quotient Rule, we can say that f(x)=sin(x)f(x) = sin(x) and g(x)=1+cos(x)g(x) = 1 + cos(x). For this specific problem, you would need to recall that…

ddx(sin(x))=cos(x)\frac{d}{dx}(sin(x)) = cos(x)

ddx(cos(x))=−sin(x)\frac{d}{dx}(cos(x)) = -sin(x)

For similar problems like this, it would be best to memorize the derivatives of all the trigonometric functions. Check out this guide if you need a review!

In doing so, we can rewrite this as:

dydx[sin(x)1+cos(x)]=1+cos(x)∗ddx(sin(x))+sin(x)∗ddx(1+cos(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{1+cos(x)*\frac{d}{dx}(sin(x)) + sin(x)*\frac{d}{dx}(1+cos(x))}{(1+cos(x))^2}

Therefore,

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x))+sin(x)(−sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)) + sin(x)(-sin(x))}{(1+cos(x))^2}

After simplifying, the final answer would be:

dydx[sin(x)1+cos(x)]=sin(x)(1+cos(x)−sin(x))(1+cos(x))2\frac{dy}{dx}\Bigg[\frac{sin(x)}{1+cos(x)}\Bigg] = \frac{sin(x)(1+cos(x)-sin(x))}{(1+cos(x))^2}

💫 Closing

Great work! Using the Quotient Rule, finding the derivative of rational functions becomes a lot easier. 🎉

Here’s a quick summary for you!

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Image of quotient rule in two forms, courtesy of OnLine Math Learning

Frequently Asked Questions

What's the formula for the quotient rule?

If you have two differentiable functions f(x) and g(x) (with g(x) ≠ 0), the quotient rule says (f/g)'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]^2. In Leibniz form, for u(x)/v(x): d/dx [u/v] = (u'v − uv')/v^2. Remember the order in the numerator matters—“low d high minus high d low” (low = denominator, high = numerator). Also note the domain restriction: g(x) must be nonzero where you apply the rule. After differentiating, simplify the rational expression when possible. This is part of AP CED FUN-3.B (derivatives of quotients). For worked examples and quick tips, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc). For more practice (1000+ problems), check the practice page (https://library.fiveable.me/practice/ap-calculus).

How do I find the derivative of x^2 divided by (x+1)?

Use the quotient rule: if f(x) = g(x)/h(x) then f'(x) = (g'(x)h(x) - g(x)h'(x)) / [h(x)]^2. Here g(x)=x^2 and h(x)=x+1, so g'(x)=2x and h'(x)=1. Compute: f'(x) = (2x(x+1) - x^2·1) / (x+1)^2 = (2x^2 + 2x - x^2) / (x+1)^2 = (x^2 + 2x) / (x+1)^2 = x(x+2) / (x+1)^2. Note domain: derivative is valid for x ≠ -1 (denominator ≠ 0). On the AP exam, show the quotient-rule setup then simplify to the final form (x(x+2))/(x+1)^2. For a quick review of the quotient rule and extra practice, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and the Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2); lots more practice problems are at (https://library.fiveable.me/practice/ap-calculus).

When do I use the quotient rule vs just dividing first?

Use the quotient rule whenever you have a true quotient f(x)/g(x) that you can’t simplify algebraically first. The quotient rule (f′g − fg′)/g^2 is in the CED (FUN-3.B.2) for finding derivatives of quotients of differentiable functions; remember the denominator mustn’t be 0. But always check if you can rewrite the quotient first—if you can simplify to a product or power, use the product, power, or chain rule (often cleaner and less algebra). Examples: - If you have x^2/x, simplify to x and use the power rule (derivative = 1). - If you have (x^3)/(x^2) simplify to x and differentiate. - If you have (x^2 + 1)/(x), rewrite as x + 1/x and use power rule on each term instead of applying quotient rule. If simplification isn’t possible (e.g., (sin x)/(x^2+1)), use the quotient rule directly. For AP free-response, show algebraic simplification when possible—that’s often faster and reduces algebraic errors. For a focused review see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and more practice problems (https://library.fiveable.me/practice/ap-calculus).

Can someone explain the quotient rule step by step?

Think of a quotient h(x) = f(x)/g(x). The quotient rule gives h′(x) in terms of f, g and their derivatives—it’s just an application of FUN-3.B from the CED. Step-by-step: 1. Identify numerator f(x) and denominator g(x). Make sure g(x) ≠ 0 (domain restriction). 2. Compute f′(x) and g′(x). 3. Plug into the quotient rule: h′(x) = [f′(x)·g(x) − f(x)·g′(x)] / [g(x)]^2. (Numerator = “derivative of top times bottom minus top times derivative of bottom”; denominator = bottom squared.) 4. Simplify algebraically when possible (factor, cancel common factors if allowed). 5. Note any points excluded because g(x)=0. Quick example: if h(x) = (x^2 + 1)/(3x − 2), f = x^2+1 → f′ = 2x; g = 3x−2 → g′ = 3. So h′(x) = [2x(3x−2) − (x^2+1)(3)] /(3x−2)^2. Simplify if needed. This is tested on AP in both procedural and contextual problems; review the Topic 2.9 study guide for more examples and practice (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc). For lots of practice problems, see (https://library.fiveable.me/practice/ap-calculus).

I keep messing up the quotient rule - do I subtract the top or bottom first?

Short answer: subtract the bottom-second. The quotient rule is (f/g)' = (g·f' − f·g') / g^2. Think “low d-high minus high d-low over low squared.” That means multiply the denominator (low) by the derivative of the numerator (high) first, then subtract the numerator times the derivative of the denominator. Order matters because subtraction isn’t commutative—swapping them flips the sign. Also remember g(x) ≠ 0 (domain restriction) and simplify when possible. Quick example: if h(x) = (x^2)/(sin x), h'(x) = (sin x·2x − x^2·cos x) / (sin x)^2. For AP practice on FUN-3.B (derivatives of quotients) check the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and try problems at (https://library.fiveable.me/practice/ap-calculus).

What's the difference between the product rule and quotient rule?

Short answer: the product rule differentiates a product f(x)·g(x); the quotient rule differentiates a quotient f(x)/g(x). Use the product rule when you multiply two differentiable functions: (fg)' = f' g + f g' (Leibniz: d/dx[f(x)g(x)] = f'(x)g(x)+f(x)g'(x)). Use the quotient rule when you divide by a nonzero differentiable function: (f/g)' = (f' g − f g') / [g]^2, with the domain restriction g(x) ≠ 0. The numerator is like “derivative of top times bottom minus top times derivative of bottom,” and you square the denominator. Practical tip: rewrite a quotient as a product f·g^−1 and apply the product rule plus the chain rule: d/dx[f·g^(−1)] = f' g^(−1) + f(−1)g^(−2) g'—which gives the same quotient-rule formula. For messy rational expressions, logarithmic differentiation can be cleaner (especially for products/quotients with powers). These are part of FUN-3.B in the CED (derivatives of products and quotients). For a focused review, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc). For broader Unit 2 review or extra practice, check (https://library.fiveable.me/ap-calculus/unit-2) and the practice set (https://library.fiveable.me/practice/ap-calculus).

How do I solve derivatives when there's a fraction with variables in both parts?

When you have a function that's a fraction f(x) = u(x)/v(x) (variables in both numerator and denominator), use the Quotient Rule (FUN-3.B.2). Formula: (u/v)' = [u'v − u v'] / [v^2]. Steps to follow: 1. Identify u(x) (numerator) and v(x) (denominator). Note v(x) ≠ 0 (domain restriction). 2. Compute u'(x) and v'(x) (use product, chain, or power rules as needed). 3. Plug into [u'v − u v'] / [v^2]. 4. Simplify algebraically and cancel factors if possible; keep denominator squared so you don’t lose domain info. Quick example: d/dx[(x^2+1)/(x^3)] = [2x·x^3 − (x^2+1)·3x^2] / x^6 = [2x^4 − 3x^4 − 3x^2]/x^6 = [−x^4 − 3x^2]/x^6 = −(x^2+3)/x^4 (for x ≠ 0). For extra practice and AP-aligned review see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) or more practice problems (https://library.fiveable.me/practice/ap-calculus).

I don't understand why the quotient rule is so complicated - is there an easier way?

Short answer: it only seems complicated because of how it’s written—you can avoid memorizing the messy form by turning a quotient into a product or using logs. Two simpler approaches (both AP-appropriate for FUN-3.B): - Product+Chain view: write h(x)=f(x)/g(x)=f(x)·[g(x)]⁻¹. Then use the product rule and chain rule: h' = f'·g⁻¹ + f·(−1)g⁻2·g' which simplifies to (f'g − fg')/g². Same result, but it feels more logical. - Logarithmic differentiation: if f and g are positive, take ln(h)=ln f − ln g, differentiate: h'/h = f'/f − g'/g, then solve for h'. This is especially helpful for messy products/quotients. Remember domain restriction: g(x) ≠ 0. Practice applying both methods so you can pick the easiest one on exam problems. For review and worked examples, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and more unit practice (https://library.fiveable.me/ap-calculus/unit-2)—and do lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Do I need to simplify after using the quotient rule or can I leave it messy?

Short answer: you don’t always have to fully simplify, but for AP answers you usually should clean it up enough to be useful and correct. Why: the Quotient Rule gives (f/g)' = (f' g − f g') / g^2—that’s correct as-is. For multiple choice you can often leave it if it matches an answer. For free-response, simplify when it helps you show domain restrictions (g(x) ≠ 0), cancel common factors only when you note any removed points, and factor or combine terms if you need critical points, limits, or sign analysis. Simplifying reduces algebra errors and makes checking work easier—AP graders want correct reasoning and clear notation. Tips: cancel common factors only after stating where g(x) ≠ 0; factor the numerator to find zeros; expand if you need to combine with other terms. For more practice and examples tied to the CED learning objective FUN-3.B, check the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

When should I use the quotient rule instead of rewriting as a negative exponent?

Use the quotient rule when the denominator is best kept as a function (especially if it might be zero or you want the derivative in rational form); rewrite with a negative exponent when doing so makes the algebra simpler and you can safely apply the power and chain/product rules. Quick checklist to help you decide: - Denominator = simple power of x (like x^n): rewrite as x^(-n) and use power/chain rules—usually easiest. - Denominator = complicated function (like g(x) not a simple power) or you want f(x)/g(x) left as a fraction: use the quotient rule f′g − fg′ over g^2 (FUN-3.B.2). - Domain issues: if g(x)=0 somewhere, keep the rational form to track domain restrictions. - Higher derivatives or messy algebra: sometimes quotient rule keeps things cleaner; other times expanding to negative exponents shortens work—pick whichever reduces algebraic complexity. For more examples and AP-style practice (aligned to Topic 2.9), see the study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

How do I find the derivative of (3x-1)/(x^2+2x)?

Use the quotient rule: if f(x)=g(x)/h(x) then f'(x) = (g'(x)h(x) − g(x)h'(x)) / [h(x)]^2. Here g(x)=3x−1, h(x)=x^2+2x. g'(x)=3, h'(x)=2x+2. So f'(x) = [3(x^2+2x) − (3x−1)(2x+2)] / (x^2+2x)^2 = [3x^2+6x − (6x^2+4x−2)] / (x^2+2x)^2 = (−3x^2+2x+2) / (x^2+2x)^2. Domain note: denominator ≠ 0, so x ≠ 0, −2. This is exactly the kind of quotient-derivative work the CED lists under FUN-3.B (Topic 2.9). For extra practice and walkthroughs on the quotient rule, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and thousands of practice problems (https://library.fiveable.me/practice/ap-calculus).

What happens if the denominator is just a constant - do I still use quotient rule?

Short answer: no—you don’t need to use the full quotient rule when the denominator is a nonzero constant. If you have h(x) = f(x)/c with c ≠ 0, rewrite h(x) = (1/c)·f(x) and use the constant multiple rule: h'(x) = (1/c)·f'(x). That follows directly from basic derivative rules in the CED (FUN-3.B, derivative of quotients/products). Why that’s okay: the quotient rule would still give the same result, since applying f'(x)c − f(x)·0 over c^2 simplifies to f'(x)/c—but doing the algebra is extra work. Always note the domain restriction: c ≠ 0. For AP prep, you should simplify answers (it’s fine to mention either rule if you show correct work). For more on the quotient rule and practice, see the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and the Unit 2 overview (https://library.fiveable.me/ap-calculus/unit-2). For extra practice problems, try (https://library.fiveable.me/practice/ap-calculus).

I'm confused about the signs in the quotient rule formula - which part is negative?

The negative sign is on the second term in the numerator. The quotient rule is (f/g)' = (f'(x)·g(x) − f(x)·g'(x)) / [g(x)]^2. So the minus applies to f(x)·g'(x)—you subtract “high (f) times derivative of low (g’)” from “derivative of high (f’) times low (g)”. A common mnemonic: “low D high minus high D low, over low squared.” Note the denominator is [g(x)]^2 (never 0 in the domain), so the sign of the whole derivative depends on the numerator (f'g − f g'). For AP language: write it in Leibniz form as (u'v − u v')/v^2 and be careful with order—swapping terms flips the sign. For extra practice on quotients and exam-style problems, check the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and the practice set (https://library.fiveable.me/practice/ap-calculus).

Can you show me how to use the quotient rule on a word problem about rates?

Sure—here's a short word-problem example that uses the quotient rule (FUN-3.B style) so you can see rates in context. Problem: A tank contains A(t) ounces of solute in V(t) ounces of solution, so concentration C(t)=A(t)/V(t). Suppose A(t)=50−2t (oz) and V(t)=10+3t (oz), with t in minutes. Find dC/dt at t=2 and interpret the result. Work (Quotient Rule): C(t)=A(t)/V(t). By the quotient rule, C′(t) = [A′(t)V(t) − A(t)V′(t)] / [V(t)]^2. Here A′(t)=−2, V′(t)=3. At t=2: A(2)=46, V(2)=16. Numerator = (−2)(16) − (46)(3) = −32 − 138 = −170. So C′(2) = −170 / 16^2 = −170 / 256 = −85/128 ≈ −0.664 (units: ounces of solute per ounce of solution per minute). Interpretation: At t=2 min the concentration is decreasing at about 0.664 oz solute per oz solution per minute. Note domain restriction V(t) ≠ 0 (here V>0). This directly matches AP curriculum keywords: Quotient Rule, numerator/denominator, domain restrictions, and FUN-3.B. For more examples and practice, check the Topic 2.9 study guide (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and lots of practice problems (https://library.fiveable.me/practice/ap-calculus).

Why does my calculator give a different answer when I use the quotient rule?

A calculator can show a “different” answer for a few normal reasons—usually those are algebraic, not wrong math. Common causes - You used numerical differentiation (nDeriv) instead of the symbolic derivative the quotient rule gives; nDeriv returns an approximate number, not a simplified formula. - Parentheses/entry error: f(x)/g(x) entered without parentheses changes g'(x) or the whole expression. - You applied the quotient rule but swapped the order in the numerator (it must be f'(x)g(x) − f(x)g'(x)). - You simplified (or cancelled) a factor after differentiating; the calculator might show an unsimplified but equivalent expression. - Domain/cancellation: if g(x)=0 at some x, algebraic cancellation before/after differentiating can change where the derivative is defined. - Symbolic simplification differences: CAS gives an equivalent expression that looks different (but is correct). What to do - Check your entry (use parentheses). - Compare: compute the symbolic derivative, simplify your result, then evaluate both at a few x-values to confirm they match numerically. - If unsure, rewrite f/g as f · g^(−1) and use product+chain rules; that can avoid sign mistakes. - Remember AP expectations: show the correct quotient-rule setup and simplify, and note domain restrictions (g(x) ≠ 0)—see the Topic 2.9 study guide for examples (https://library.fiveable.me/ap-calculus/unit-2/quotient-rule/study-guide/qjB4zQXXy6ps14FlOXLc) and more practice at (https://library.fiveable.me/practice/ap-calculus).