Quotient Rule — JEE Main Basic Maths MCQs with Solutions
Free JEE Main Basic Maths Quotient Rule MCQs with step-by-step solutions (30 questions). Part of Differentiation. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Quotient Rule · easy · theory
The quotient rule for y = u(x)/v(x) states dy/dx =
A. $\dfrac{v\,u' - u\,v'}{v^2}$ ✓ Correct
B. $\dfrac{u'}{v'}$
C. $\dfrac{v\,u' - u\,v'}{v}$
D. $\dfrac{u\,v' - v\,u'}{v^2}$
Solution: Quotient rule: (u/v)′ = (v u′ − u v′)/v² — "denominator times derivative of numerator, minus numerator times derivative of denominator, all over denominator squared."
Q2 — Quotient Rule · easy · theory
In the quotient rule numerator (v u′ − u v′), the ORDER of the two terms:
A. Matters — reversing it changes the sign of the whole derivative ✓ Correct
B. Only matters for trigonometric functions
C. Does not matter
D. Is always symmetric
Solution: v u′ − u v′ ≠ u v′ − v u′; swapping them negates the answer. This wrong order is the most frequent quotient-rule error.
Q3 — Quotient Rule · easy · theory
In the quotient rule the denominator of the result is:
A. v² (the denominator squared) ✓ Correct
B. 2v
C. v'²
D. v
Solution: The result is divided by v², the square of the original denominator. Forgetting to square it is a common slip.
Q4 — Quotient Rule · easy · theory
The derivative of y = 1/v(x) (reciprocal rule) is:
A. $\dfrac{v'(x)}{[v(x)]^2}$
B. $-\dfrac{1}{[v(x)]^2}$
C. $\dfrac{1}{v'(x)}$
D. $-\dfrac{v'(x)}{[v(x)]^2}$ ✓ Correct
Solution: Taking u = 1 (u′ = 0) in the quotient rule: (1/v)′ = (v·0 − 1·v′)/v² = −v′/v².
Q5 — Quotient Rule · medium · theory
The quotient rule can always be avoided by rewriting u/v as:
A. The product u·v⁻¹ and applying the product rule (with a chain-rule step) ✓ Correct
B. The sum u + v
C. A single power of x
D. The difference u − v
Solution: u/v = u·v⁻¹; the product rule with d/dx(v⁻¹) = −v⁻²v′ reproduces the quotient rule.
Q6 — Quotient Rule · easy · theory
Which function is most simply differentiated WITHOUT the quotient rule?
A. $y = \dfrac{\sin x}{x^2}$
B. $y = \dfrac{x^3}{x} = x^2$ ✓ Correct
C. $y = \dfrac{x^2+1}{x-3}$
D. $y = \dfrac{e^x}{1+e^x}$
Solution: x³/x simplifies to x² (for x ≠ 0), giving 2x directly. Always simplify before reaching for the quotient rule.
Q7 — Quotient Rule · medium · theory
Applying the quotient rule to y = sin x / cos x gives:
A. $\tan x$
B. $-\csc^2 x$
C. $1$
D. $\sec^2 x$ ✓ Correct
Solution: (cos x·cos x − sin x·(−sin x))/cos²x = (cos²x + sin²x)/cos²x = 1/cos²x = sec²x. This proves d/dx(tan x) = sec²x.
Q8 — Quotient Rule · easy · theory
The derivative of y = c/v(x), where c is a constant, is:
A. $-\dfrac{c\,v'}{v^2}$ ✓ Correct
B. $0$
C. $\dfrac{c}{v'}$
D. $\dfrac{c\,v'}{v^2}$
Solution: u = c (u′ = 0): (c/v)′ = (v·0 − c v′)/v² = −c v′/v².
Q9 — Quotient Rule · easy · theory
For y = u/v with u and v both functions of x, the quotient rule is valid provided:
A. u and v are polynomials only
B. u and v are differentiable and v(x) ≠ 0 ✓ Correct
C. u(x) = 0
D. v(x) = 1
Solution: The quotient rule applies to any differentiable u and v wherever the denominator v is non-zero.
Q10 — Quotient Rule · easy · theory
For y = (ln x)/x, identifying u and v gives:
A. u = ln x, v = 1
B. u = ln x, v = x ✓ Correct
C. u = 1, v = x ln x
D. u = x, v = ln x
Solution: The numerator is u = ln x and the denominator is v = x, so y′ = (x·(1/x) − ln x·1)/x² = (1 − ln x)/x².
Q11 — Quotient Rule · medium · theory
When both numerator and denominator are composite (e.g. y = sin(2x)/e^{3x}), differentiating requires:
A. Only the chain rule
B. Only the product rule
C. The quotient rule together with the chain rule on each part ✓ Correct
D. No rule at all
Solution: Each of u′ and v′ needs the chain rule (2cos 2x and 3e^{3x}), fed into the quotient rule.
Q12 — Quotient Rule · easy · theory
A neat way to remember the quotient-rule numerator order is:
A. "d-High times d-Low"
B. "Low d-High minus High d-Low, over Low squared" ✓ Correct
C. "High d-Low minus Low d-High, over Low squared"
D. "High plus Low, over Low"
Solution: "Lo·dHi − Hi·dLo over Lo²": denominator (Lo) times derivative of numerator (Hi) first, keeping the correct sign.
Q13 — Quotient Rule · medium · numerical
If $y = \dfrac{x^2+1}{x-3}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{2x}{1}$
B. $\dfrac{x^2+6x+1}{(x-3)^2}$
C. $\dfrac{-x^2-6x-1}{(x-3)^2}$
D. $\dfrac{x^2-6x-1}{(x-3)^2}$ ✓ Correct
Solution: [(x−3)(2x) − (x²+1)(1)]/(x−3)² = (2x²−6x−x²−1)/(x−3)² = (x²−6x−1)/(x−3)².
Q14 — Quotient Rule · medium · numerical
If $y = \dfrac{\sin x}{x^2}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{x\cos x - 2\sin x}{x^3}$ ✓ Correct
B. $\dfrac{x\cos x + 2\sin x}{x^3}$
C. $\dfrac{\cos x}{2x}$
D. $\dfrac{\cos x - 2\sin x}{x^3}$
Solution: [x²cos x − sin x·2x]/x⁴ = [x cos x − 2 sin x]/x³.
Q15 — Quotient Rule · medium · numerical
If $y = \dfrac{e^x}{1+e^x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{-e^x}{(1+e^x)^2}$
B. $\dfrac{e^x}{1+e^x}$
C. $\dfrac{e^x}{(1+e^x)^2}$ ✓ Correct
D. $\dfrac{e^{2x}}{(1+e^x)^2}$
Solution: [(1+eˣ)eˣ − eˣ·eˣ]/(1+eˣ)² = eˣ/(1+eˣ)².
Q16 — Quotient Rule · easy · numerical
If $y = \dfrac{\ln x}{x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{1-\ln x}{x^2}$ ✓ Correct
B. $\dfrac{1+\ln x}{x^2}$
C. $\dfrac{1}{x^2}$
D. $\dfrac{\ln x - 1}{x^2}$
Solution: [x·(1/x) − ln x·1]/x² = (1 − ln x)/x².
Q17 — Quotient Rule · easy · numerical
If $y = \dfrac{x}{x+1}$, then $\dfrac{dy}{dx}$ =
A. $1$
B. $\dfrac{1}{(x+1)^2}$ ✓ Correct
C. $\dfrac{-1}{(x+1)^2}$
D. $\dfrac{2x+1}{(x+1)^2}$
Solution: [(x+1)(1) − x(1)]/(x+1)² = 1/(x+1)².
Q18 — Quotient Rule · easy · numerical
If $y = \dfrac{1-x}{1+x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{2}{(1+x)^2}$
B. $\dfrac{-2}{(1+x)^2}$ ✓ Correct
C. $\dfrac{-1}{(1+x)^2}$
D. $\dfrac{-2x}{(1+x)^2}$
Solution: [(1+x)(−1) − (1−x)(1)]/(1+x)² = (−1−x−1+x)/(1+x)² = −2/(1+x)².
Q19 — Quotient Rule · easy · numerical
If $y = \dfrac{2x+3}{x-1}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{5}{(x-1)^2}$
B. $\dfrac{-1}{(x-1)^2}$
C. $\dfrac{2}{(x-1)^2}$
D. $\dfrac{-5}{(x-1)^2}$ ✓ Correct
Solution: [(x−1)(2) − (2x+3)(1)]/(x−1)² = (2x−2−2x−3)/(x−1)² = −5/(x−1)².
Q20 — Quotient Rule · medium · numerical
If $y = \dfrac{\cos x}{1+\sin x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{-1}{1+\sin x}$ ✓ Correct
B. $\dfrac{1}{1+\sin x}$
C. $\dfrac{1}{(1+\sin x)^2}$
D. $\dfrac{-\cos x}{(1+\sin x)^2}$
Solution: [(1+sin x)(−sin x) − cos x·cos x]/(1+sin x)² = [−sin x − (sin²x+cos²x)]/(1+sin x)² = −(1+sin x)/(1+sin x)² = −1/(1+sin x).
Q21 — Quotient Rule · easy · numerical
If $y = \dfrac{x^2}{e^x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{2x-x^2}{e^x}$ ✓ Correct
B. $\dfrac{2x}{e^x}$
C. $\dfrac{x^2+2x}{e^x}$
D. $\dfrac{x^2-2x}{e^x}$
Solution: [eˣ·2x − x²eˣ]/e^{2x} = (2x − x²)/eˣ = x(2 − x)e^{−x}.
Q22 — Quotient Rule · easy · numerical
If $y = \dfrac{e^x}{x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{e^x(x+1)}{x^2}$
B. $\dfrac{e^x(1-x)}{x^2}$
C. $\dfrac{e^x}{x^2}$
D. $\dfrac{e^x(x-1)}{x^2}$ ✓ Correct
Solution: [x·eˣ − eˣ·1]/x² = eˣ(x−1)/x².
Q23 — Quotient Rule · medium · numerical
If $y = \dfrac{x}{\ln x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{1}{(\ln x)^2}$
B. $\dfrac{\ln x - 1}{(\ln x)^2}$ ✓ Correct
C. $\dfrac{1-\ln x}{(\ln x)^2}$
D. $\dfrac{\ln x + 1}{(\ln x)^2}$
Solution: [ln x·1 − x·(1/x)]/(ln x)² = (ln x − 1)/(ln x)².
Q24 — Quotient Rule · medium · numerical
If $y = \dfrac{x^2-1}{x^2+1}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{2x}{(x^2+1)^2}$
B. $\dfrac{4x^3}{(x^2+1)^2}$
C. $\dfrac{-4x}{(x^2+1)^2}$
D. $\dfrac{4x}{(x^2+1)^2}$ ✓ Correct
Solution: [(x²+1)(2x) − (x²−1)(2x)]/(x²+1)² = 2x[(x²+1) − (x²−1)]/(x²+1)² = 4x/(x²+1)².
Q25 — Quotient Rule · easy · numerical
If $y = \dfrac{1}{x^2+1}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{-2x}{x^2+1}$
B. $\dfrac{2x}{(x^2+1)^2}$
C. $\dfrac{-1}{(x^2+1)^2}$
D. $\dfrac{-2x}{(x^2+1)^2}$ ✓ Correct
Solution: Reciprocal rule: −v′/v² = −2x/(x²+1)².
Q26 — Quotient Rule · medium · numerical
If $y = \dfrac{e^x-1}{e^x+1}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{e^x}{(e^x+1)^2}$
B. $\dfrac{2e^x}{(e^x+1)^2}$ ✓ Correct
C. $\dfrac{-2e^x}{(e^x+1)^2}$
D. $\dfrac{2e^{2x}}{(e^x+1)^2}$
Solution: [(eˣ+1)eˣ − (eˣ−1)eˣ]/(eˣ+1)² = eˣ[(eˣ+1) − (eˣ−1)]/(eˣ+1)² = 2eˣ/(eˣ+1)².
Q27 — Quotient Rule · easy · numerical
If $y = \dfrac{\cos x}{x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{-(x\sin x + \cos x)}{x^2}$ ✓ Correct
B. $\dfrac{\cos x - x\sin x}{x^2}$
C. $\dfrac{-\sin x}{x}$
D. $\dfrac{x\sin x - \cos x}{x^2}$
Solution: [x(−sin x) − cos x(1)]/x² = −(x sin x + cos x)/x².
Q28 — Quotient Rule · medium · numerical
For $y = \dfrac{x}{x^2+1}$, the value of $\dfrac{dy}{dx}$ at $x = 0$ is:
A. 0
B. −1
C. 2
D. 1 ✓ Correct
Solution: y′ = [(x²+1)(1) − x(2x)]/(x²+1)² = (1 − x²)/(x²+1)²; at x = 0: (1−0)/1 = 1.
Q29 — Quotient Rule · easy · numerical
If $y = \dfrac{1+x^2}{x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{x^2+1}{x^2}$
B. $2$
C. $\dfrac{1-x^2}{x^2}$
D. $\dfrac{x^2-1}{x^2}$ ✓ Correct
Solution: [x·2x − (1+x²)(1)]/x² = (2x² − 1 − x²)/x² = (x² − 1)/x². (Or write y = x⁻¹ + x ⇒ y′ = −x⁻² + 1.)
Q30 — Quotient Rule · easy · numerical
If $y = \dfrac{3x}{x^2+1}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{3(x^2-1)}{(x^2+1)^2}$
B. $\dfrac{3}{(x^2+1)^2}$
C. $\dfrac{3(1-x^2)}{(x^2+1)^2}$ ✓ Correct
D. $\dfrac{6x}{(x^2+1)^2}$
Solution: [(x²+1)(3) − 3x(2x)]/(x²+1)² = (3x²+3−6x²)/(x²+1)² = 3(1−x²)/(x²+1)².