📝 Quotient rule for derivatives (17 MCQs)

📖 From Calculus • 3. The Derivation • 17 questions available

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What is Quotient rule for derivatives?

Definition:
The quotient rule states that the derivative of a quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

Example:
For $f(x) = \frac{x}{x+1}$ , the derivative is $f'(x) = \frac{(x+1)(1) - x(1)}{(x+1)^2} = \frac{1}{(x+1)^2}$ .

Reason:
This rule handles rational functions systematically, avoiding mistakes from incorrect application of other rules, and is vital for differentiating fractions involving variables in both numerator and denominator positions.

Easy

Medium

Hard

📝 All Quotient rule for derivatives MCQs

Q1. What is the derivative of the quotient $\frac{f(x)}{g(x)}$ expressed in terms of $f, g$ and their derivatives?

\displaystyle \frac{f&#039;(x)}{g&#039;(x)}

\displaystyle \frac{g&#039;(x)}{f&#039;(x)}

\displaystyle \frac{g(x)f&#039;(x)-f(x)g&#039;(x)}{[g(x)]^{2}}

✅

\displaystyle \frac{f(x)g&#039;(x)-g(x)f&#039;(x)}{[f(x)]^{2}}

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Applying the quotient rule gives $\frac{d}{dx}\bigl(\frac{f}{g}\bigr)=\frac{g\,f'-f\,g'}{g^{2}}$ . This matches option C, while the other options either invert the fraction or misuse the denominator, so C is the correct formula.

Q2. For the quotient rule to be valid at a point $x=a$ , which of the following must be true about $g(a)$ ?

g(a)=0

g(a)\neq0

✅

g(a)>0

g(a)<0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The rule requires division by $g(x)$ ; if $g(a)=0$ the expression would be undefined. Therefore the only acceptable condition is that $g(a)$ be non‑zero, making option B the correct choice.

Q3. Suppose $f(x)=x^{2}$ and $g(x)=\sin x$ . At $x=\pi/2$ , which statement about the sign of $\frac{d}{dx}\bigl(\frac{f}{g}\bigr)$ is correct?

A.Positive ✅

B.Negative

C.Zero

D.Cannot be determined without further computation

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using the quotient rule, $(g f' - f g')/g^{2} = (1\cdot 2x - x^{2}\cdot0)/1^{2}$ at $x=\pi/2$ yields a positive numerator, so the derivative is positive. Hence option A is correct.

Q4. Which approach yields a simpler derivative for $h(x)=\frac{x^{3}}{x}$ ?

A.Apply the quotient rule directly

B.Simplify to

x^{2}

before differentiating ✅

C.Use the product rule after rewriting as

x^{3}\cdot x^{-1}

D.Differentiate numerator and denominator separately then divide

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The expression simplifies to $x^{2}$ algebraically; differentiating this gives $2x$ instantly. Directly applying the quotient rule would create unnecessary algebra, so simplifying first (option B) is the most efficient.

Q5. If $f$ and $g$ are differentiable and $g(x)=e^{x}$ , how does the presence of $e^{x}$ in the denominator affect the form of the quotient rule result compared to a generic $g(x)$ ?

A.The denominator squared becomes

e^{2x}

and

g&#039;(x)=e^{x}

simplifies the numerator. ✅

B.The denominator disappears because

e^{x}

cancels.

C.The rule cannot be applied when

g(x)=e^{x}

D.The derivative reduces to

f&#039;(x)/e^{x}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: When $g(x)=e^{x}$ , its derivative is also $e^{x}$ . Substituting into the quotient rule gives $(e^{x}f' - f e^{x})/(e^{x})^{2} = (e^{x}(f'-f))/e^{2x}$ , which matches option A. The other statements misrepresent the algebraic outcome.

Q6. Let $p(x)=\frac{u(x)}{v(x)}$ where $u$ and $v$ are twice differentiable. If $p'(a)=0$ and $v(a)\neq0$ , which of the following must be true?

u&#039;(a)=0

u(a)=0

u&#039;(a)v(a)=u(a)v&#039;(a)

✅

v&#039;(a)=0

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Setting the quotient derivative $(v u' - u v')/v^{2}=0$ forces the numerator $v u' - u v'$ to vanish, giving the relationship $u'(a)v(a)=u(a)v'(a)$ . This is precisely option C.

Q7. You have $q(x)=\frac{x^{2}+1}{x-1}$ . Which of the following expressions represents $q'(x)$ after applying the quotient rule and simplifying?

\displaystyle \frac{(x-1)(2x)-(x^{2}+1)(1)}{(x-1)^{2}}

\displaystyle \frac{2x(x-1)-(x^{2}+1)}{(x-1)^{2}}

✅

\displaystyle \frac{2x(x-1)+(x^{2}+1)}{(x-1)^{2}}

\displaystyle \frac{2x(x-1)-(x^{2}+1)}{(x-1)}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using the rule gives $(x-1)(2x)-(x^{2}+1)(1)$ over $(x-1)^{2}$ . Factoring the numerator yields $2x(x-1)-(x^{2}+1)$ , which is exactly option B.

Q8. Consider the function $r(x)=\frac{\ln(x)}{x^{2}}$ . Which of the following best describes the behavior of $r'(x)$ as $x\to\infty$ ?

A.It approaches 0 from the positive side.

B.It approaches 0 from the negative side. ✅

C.It grows without bound.

D.It oscillates.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Differentiating gives $r'(x)=\frac{1-2\ln x}{x^{3}}$ . As $x$ grows, $\ln x$ dominates, making the numerator negative while the denominator stays positive, so the derivative tends to 0 from below, matching option B.

Q9. Given $s(x)=\frac{e^{2x}}{x^{3}}$ . If $s'(c)=0$ for some positive $c$ , which equation must $c$ satisfy?

2e^{2c}x^{3}-3e^{2c}x^{2}=0

2c-3=0

2c=3

✅

2c^{3}-3c^{2}=0

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Applying the rule yields $s'(x)=e^{2x}(2x-3)/x^{4}$ . Setting this equal to zero forces the factor $2x-3$ to vanish, giving $2c-3=0$ or $c=3/2$ . Hence option C is correct.

Q10. Which of the following functions, after applying the quotient rule, will result in a derivative that simplifies to a constant (non‑zero) expression?

\displaystyle \frac{x}{x}

\displaystyle \frac{x^{2}}{x}

✅

\displaystyle \frac{x}{x^{2}}

\displaystyle \frac{x^{3}}{x^{2}}

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $f(x)=x^{2}$ and $g(x)=x$ , the quotient rule gives $(x\cdot2x-x^{2}\cdot1)/x^{2}=1$ , a constant non‑zero derivative. The other choices either simplify to a constant zero or produce a variable expression, so B is the right answer.

Q11. When simplifying a quotient before differentiating, which principle justifies that the derivative of the simplified form equals the derivative obtained via the quotient rule?

A.The chain rule

B.Algebraic equivalence of functions ✅

C.L'Hôpital's rule

D.Mean value theorem

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: If two expressions are algebraically identical for all $x$ in their domain, they represent the same function. Differentiating either form must yield the same derivative, so the justification is algebraic equivalence, option B.

Q12. If $f(x)=\sqrt{x}$ and $g(x)=x$ , what is the sign of $\frac{d}{dx}\bigl(\frac{f}{g}\bigr)$ for $x>0$ ?

A.Positive

B.Negative ✅

C.Zero

D.Undefined

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Using the quotient rule gives $\frac{x\cdot\frac{1}{2\sqrt{x}}-\sqrt{x}\cdot1}{x^{2}}=-\frac{1}{2x^{3/2}}$ , which is negative for all positive $x$ . Therefore option B is correct.

Q13. For the function $t(x)=\frac{x^{4}-1}{x^{2}+1}$ , which of the following is the correct expression for $t'(x)$ after full simplification?

\displaystyle \frac{4x^{3}(x^{2}+1)-(x^{4}-1)2x}{(x^{2}+1)^{2}}

\displaystyle \frac{4x^{3}(x^{2}+1)-2x(x^{4}-1)}{(x^{2}+1)^{2}}

✅

\displaystyle \frac{4x^{3}-2x}{(x^{2}+1)}

\displaystyle \frac{4x^{3}-2x}{(x^{2}+1)^{2}}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Applying the quotient rule yields $(x^{2}+1)(4x^{3})-(x^{4}-1)(2x)$ over $(x^{2}+1)^{2}$ . This matches option B after factoring the common $2x$ term, confirming it as the correct simplified derivative.

Q14. Why does the quotient rule contain the denominator squared in its final denominator?

A.Because the derivative of a product introduces an extra factor of the denominator.

B.It results from applying the product rule to

f(x)\cdot[g(x)]^{-1}

. ✅

C.The square ensures the units match.

D.It is a mistake; the denominator should be first power.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Writing the quotient as $f\cdot g^{-1}$ and applying the product rule gives $f' g^{-1}+f(-g^{-2}g')$ . Combining terms over the common denominator $g^{2}$ yields the squared denominator, which is exactly the reasoning in option B.

Q15. Let $u(x)=\ln(x)$ and $v(x)=x^{k}$ where $k\neq0$ . For which value(s) of $k$ does the derivative of $\frac{u}{v}$ become zero at exactly one positive $x$ value?

A.All

k>0

B.All

k<0

C.All

k\neq0

✅

D.Only

k=1

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Setting $(v\,u' - u\,v')=0$ gives $1 - k\ln x =0$ → $\ln x = 1/k$ . For any non‑zero $k$ there is exactly one positive solution $x=e^{1/k}$ . Thus the condition holds for all $k\neq0$ , making option C correct.

Q16. If $w(x)=\frac{3x^{2}+2}{x-4}$ , what is $w'(2)$ ?

-\frac{38}{4}

✅

-\frac{22}{4}

\frac{38}{4}

\frac{22}{4}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Using the quotient rule: $w'=\frac{(x-4)(6x)-(3x^{2}+2)}{(x-4)^{2}}$ . Substituting $x=2$ gives $\frac{(-2)(12)-(12+2)}{4}= \frac{-24-14}{4}= -\frac{38}{4}$ . Hence option A is correct.

Q17. When differentiating $y=\frac{f(x)}{g(x)}$ where both $f$ and $g$ have critical points at $x=c$ , which of the following statements is always true about $y'(c)$ ?

y&#039;(c)=0

✅

y&#039;(c)

is undefined

y&#039;(c)=\frac{f&#039;(c)}{g&#039;(c)}

y&#039;(c)

depends on the relative sizes of

f&#039;(c)

and

g&#039;(c)

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: If both $f'(c)=0$ and $g'(c)=0$ , the numerator of the quotient rule becomes $g(c)\cdot0 - f(c)\cdot0 =0$ . Provided $g(c)\neq0$ , the denominator is non‑zero, so $y'(c)=0$ . Hence option A holds universally.

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