Product Rule — MH-CET Basic Maths MCQs with Solutions
Free MH-CET Basic Maths Product 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 — Product Rule · easy · theory
The product rule for y = u(x)·v(x) states dy/dx =
A. $u\,v' - v\,u'$
B. $u\,v' + v\,u'$ ✓ Correct
C. $\dfrac{u'v - uv'}{v^2}$
D. $u'\,v'$
Solution: Product rule: (uv)′ = u v′ + v u′ — "first times derivative of second, plus second times derivative of first."
Q2 — Product Rule · easy · theory
Which expression genuinely requires the PRODUCT rule (not just scalar multiplication)?
A. $y = x^2 \sin x$ ✓ Correct
B. $y = \pi e^x$
C. $y = 3x^2$
D. $y = 5\sin x$
Solution: x²·sin x is a product of two functions of x. The others are a constant times a single function, needing only the constant-multiple rule.
Q3 — Product Rule · easy · theory
For y = x·eˣ, choosing u = x and v = eˣ, the product rule gives dy/dx =
A. $x e^x$
B. $e^x$
C. $x e^x - e^x$
D. $x e^x + e^x = e^x(x+1)$ ✓ Correct
Solution: u v' + v u' = x·eˣ + eˣ·1 = eˣ(x + 1).
Q4 — Product Rule · easy · theory
A key advantage of writing a product as u·v before differentiating is that:
A. Each factor can be differentiated separately and combined by the rule ✓ Correct
B. The rule is unnecessary
C. The product becomes a sum automatically
D. The derivative is simply u′v′
Solution: The product rule lets you handle each factor with its own known derivative, then combine — you never just multiply the two derivatives.
Q5 — Product Rule · easy · theory
A common ERROR in applying the product rule is to write (uv)′ as:
A. $uv' + vu'$
B. $vu' + uv'$
C. $u'v'$ (product of the derivatives) ✓ Correct
D. the correct combination
Solution: Writing (uv)′ = u′v′ is wrong; the correct rule has two terms: uv′ + vu′.
Q6 — Product Rule · medium · theory
To differentiate y = x²·sin x·eˣ (a product of THREE functions), the extended product rule gives:
A. $uvw$
B. $u'vw$ only
C. $u'v'w'$
D. $u'vw + uv'w + uvw'$ ✓ Correct
Solution: For three factors: (uvw)′ = u′vw + uv′w + uvw′ — differentiate one factor at a time, keeping the others.
Q7 — Product Rule · medium · theory
When differentiating y = e^{2x}·cos 3x, besides the product rule you also need:
A. No additional rule
B. Only the quotient rule
C. Integration by parts
D. The chain rule for both e^{2x} and cos 3x ✓ Correct
Solution: Both factors are composites, so each factor's derivative (2e^{2x} and −3sin 3x) requires the chain rule inside the product rule.
Q8 — Product Rule · easy · theory
The product rule can be derived from the definition of the derivative and is valid for:
A. Only trigonometric functions
B. Only polynomial functions
C. Any two differentiable functions ✓ Correct
D. Only when one factor is a constant
Solution: The product rule holds for the product of any two differentiable functions, of any type.
Q9 — Product Rule · easy · theory
For y = (x+1)(x+2), one may either use the product rule OR:
A. It cannot be differentiated another way
B. Use the quotient rule
C. Expand to x² + 3x + 2 and differentiate term by term ✓ Correct
D. Use the chain rule
Solution: Both give 2x + 3. Expanding first is a valid shortcut for simple polynomial products.
Q10 — Product Rule · easy · theory
In the product rule (uv)′ = uv′ + vu′, the two terms are:
A. Subtracted
B. Divided
C. Multiplied together
D. Added, never subtracted ✓ Correct
Solution: The product rule always ADDS the two terms. (Subtraction appears in the quotient rule, a common point of confusion.)
Q11 — Product Rule · easy · theory
To differentiate y = x·ln x, the required derivatives of the factors are:
A. $u' = x$ and $v' = 1$
B. $u' = 0$ and $v' = 1/x$
C. $u' = 1$ (for x) and $v' = 1/x$ (for ln x) ✓ Correct
D. $u' = 1$ and $v' = \ln x$
Solution: For u = x, u′ = 1; for v = ln x, v′ = 1/x. Then y′ = x(1/x) + ln x·1 = 1 + ln x.
Q12 — Product Rule · medium · theory
The product rule reduces to the constant-multiple rule when:
A. The product is zero
B. Both factors are constants
C. One of the two factors is a constant (its derivative is zero) ✓ Correct
D. Both factors are equal
Solution: If v = c (constant), v′ = 0, so (uc)′ = u·0 + c·u′ = c u′ — the constant-multiple rule.
Q13 — Product Rule · easy · numerical
If $y = x^2 \sin x$, then $\dfrac{dy}{dx}$ =
A. $2x\cos x$
B. $x^2\cos x - 2x\sin x$
C. $x^2\cos x + 2x\sin x$ ✓ Correct
D. $2x\sin x\cos x$
Solution: u = x², v = sin x: y′ = x²cos x + sin x·2x = x²cos x + 2x sin x.
Q14 — Product Rule · easy · numerical
If $y = e^x \ln x$, then $\dfrac{dy}{dx}$ =
A. $e^x\left(\ln x - \dfrac{1}{x}\right)$
B. $e^x\ln x$
C. $\dfrac{e^x}{x}$
D. $e^x\ln x + \dfrac{e^x}{x}$ ✓ Correct
Solution: y′ = eˣ(1/x) + ln x·eˣ = eˣ ln x + eˣ/x = eˣ(ln x + 1/x).
Q15 — Product Rule · easy · numerical
If $y = x\cos x$, then $\dfrac{dy}{dx}$ =
A. $-x\sin x$
B. $-\sin x$
C. $\cos x + x\sin x$
D. $\cos x - x\sin x$ ✓ Correct
Solution: y′ = x(−sin x) + cos x·1 = cos x − x sin x.
Q16 — Product Rule · easy · numerical
If $y = x^3 e^x$, then $\dfrac{dy}{dx}$ =
A. $x^3 e^x + 3x^2 e^x = x^2 e^x(x+3)$ ✓ Correct
B. $3x^2 e^x$
C. $x^2 e^x(x-3)$
D. $x^3 e^x$
Solution: y′ = x³eˣ + eˣ·3x² = x²eˣ(x + 3).
Q17 — Product Rule · medium · numerical
If $y = \sqrt{x}\,\sin x$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{\sin x}{2\sqrt{x}}$
B. $\sqrt{x}\cos x + \dfrac{\sin x}{2\sqrt{x}}$ ✓ Correct
C. $\sqrt{x}\cos x$
D. $\dfrac{\cos x}{2\sqrt{x}}$
Solution: u = √x (u′ = 1/(2√x)), v = sin x: y′ = √x cos x + sin x/(2√x).
Q18 — Product Rule · easy · numerical
If $y = x\tan x$, then $\dfrac{dy}{dx}$ =
A. $x\sec^2 x$
B. $\tan x + x\sec^2 x$ ✓ Correct
C. $\tan x - x\sec^2 x$
D. $\sec^2 x$
Solution: y′ = x·sec²x + tan x·1 = tan x + x sec²x.
Q19 — Product Rule · medium · numerical
If $y = e^{2x}\cos 3x$, then $\dfrac{dy}{dx}$ =
A. $e^{2x}(2\cos 3x + 3\sin 3x)$
B. $2e^{2x}\cos 3x$
C. $-6e^{2x}\sin 3x$
D. $e^{2x}(2\cos 3x - 3\sin 3x)$ ✓ Correct
Solution: y′ = e^{2x}(−3sin 3x) + cos 3x·2e^{2x} = e^{2x}(2cos 3x − 3sin 3x). (Product + chain rule on both factors.)
Q20 — Product Rule · medium · numerical
If $y = (x^2+1)(x^3-2)$, then $\dfrac{dy}{dx}$ =
A. $6x^3 - 4x$
B. $(2x)(3x^2)$
C. $5x^4 + 3x^2 - 4x$ ✓ Correct
D. $5x^4 + 3x^2 + 4x$
Solution: y′ = (x²+1)(3x²) + (x³−2)(2x) = 3x⁴ + 3x² + 2x⁴ − 4x = 5x⁴ + 3x² − 4x.
Q21 — Product Rule · medium · numerical
If $y = x^2 e^x$, its derivative at $x = 0$ is:
A. 1
B. 0 ✓ Correct
C. e
D. 2
Solution: y′ = x²eˣ + 2x eˣ = xeˣ(x + 2); at x = 0: 0·e⁰·2 = 0.
Q22 — Product Rule · medium · numerical
If $y = \sin x\cos x$, then $\dfrac{dy}{dx}$ =
A. $-\sin 2x$
B. $\sin 2x$
C. $-\cos 2x$
D. $\cos 2x$ ✓ Correct
Solution: y′ = sin x(−sin x) + cos x·cos x = cos²x − sin²x = cos 2x. (Or note y = ½sin 2x ⇒ y′ = cos 2x.)
Q23 — Product Rule · easy · numerical
If $y = x\ln x$, then $\dfrac{dy}{dx}$ =
A. $\ln x$
B. $1 - \ln x$
C. $1 + \ln x$ ✓ Correct
D. $\dfrac{1}{x}$
Solution: y′ = x(1/x) + ln x·1 = 1 + ln x.
Q24 — Product Rule · easy · numerical
If $y = (2x+1)e^x$, then $\dfrac{dy}{dx}$ =
A. $2e^x$
B. $e^x(2x+3)$ ✓ Correct
C. $e^x(2x+1)$
D. $e^x(2x-1)$
Solution: y′ = (2x+1)eˣ + eˣ·2 = eˣ(2x + 3).
Q25 — Product Rule · medium · numerical
If $y = x^2\ln x$, then $\dfrac{dy}{dx}$ =
A. $2x\ln x$
B. $x(2\ln x + 1)$ ✓ Correct
C. $x(2\ln x - 1)$
D. $x\ln x$
Solution: y′ = x²(1/x) + ln x·2x = x + 2x ln x = x(1 + 2ln x).
Q26 — Product Rule · medium · numerical
If $y = e^x\sin x$, its derivative at $x = 0$ is:
A. e
B. 0
C. 1 ✓ Correct
D. 2
Solution: y′ = eˣsin x + eˣcos x = eˣ(sin x + cos x); at x = 0: 1·(0 + 1) = 1.
Q27 — Product Rule · medium · numerical
If $y = x\,e^{x^2}$, then $\dfrac{dy}{dx}$ =
A. $e^{x^2}(1 + 2x^2)$ ✓ Correct
B. $2x^2 e^{x^2}$
C. $e^{x^2}$
D. $e^{x^2}(1 - 2x^2)$
Solution: y′ = x·e^{x²}·2x + e^{x²}·1 = e^{x²}(2x² + 1). (Product rule with a chain-rule factor.)
Q28 — Product Rule · medium · numerical
If $y = (x-1)\sqrt{x}$, then $\dfrac{dy}{dx}$ =
A. $\dfrac{3x-1}{2\sqrt{x}}$ ✓ Correct
B. $\dfrac{x-1}{2\sqrt{x}}$
C. $\sqrt{x}$
D. $\dfrac{3x+1}{2\sqrt{x}}$
Solution: y′ = (x−1)/(2√x) + √x = (x−1)/(2√x) + 2x/(2√x) = (3x−1)/(2√x).
Q29 — Product Rule · medium · numerical
If $y = \sin x\ln x$, then $\dfrac{dy}{dx}$ =
A. $\cos x\ln x$
B. $\dfrac{\cos x}{x}$
C. $\cos x\ln x - \dfrac{\sin x}{x}$
D. $\cos x\ln x + \dfrac{\sin x}{x}$ ✓ Correct
Solution: y′ = sin x(1/x) + ln x·cos x = cos x ln x + (sin x)/x.
Q30 — Product Rule · easy · numerical
If $y = (3x^2)(\cos x)$, then $\dfrac{dy}{dx}$ =
A. $3x^2\sin x$
B. $-6x\sin x$
C. $6x\cos x + 3x^2\sin x$
D. $6x\cos x - 3x^2\sin x$ ✓ Correct
Solution: y′ = 3x²(−sin x) + cos x·6x = 6x cos x − 3x² sin x.