📝 Product rule for derivatives (18 MCQs)

📖 From Calculus • 3. The Derivation • 18 questions available

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What is Product rule for derivatives?

Definition:
The product rule states that the derivative of a product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first, expressed as $\frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$ .

Example:
For $f(x) = x^2 \sin(x)$ , the derivative is $f'(x) = 2x \sin(x) + x^2 \cos(x)$ , applying the rule correctly.

Reason:
This rule is essential when functions are multiplied rather than added, preventing common errors and enabling accurate differentiation of composite algebraic-trigonometric expressions frequently encountered in applied mathematics.

Easy

Medium

Hard

📝 All Product rule for derivatives MCQs

Q1. Which formula correctly states the product rule for differentiation?

(f\cdot g)&#039; = f&#039; \cdot g&#039;

(f\cdot g)&#039; = f \cdot g&#039; + g \cdot f&#039;

✅

(f\cdot g)&#039; = f&#039; \cdot g + f \cdot g&#039;

(f\cdot g)&#039; = f + g

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The product rule tells us that the derivative of a product is the first function times the derivative of the second plus the second function times the derivative of the first, which is expressed by the second option.

Q2. Identify the statement that is NOT a valid consequence of the product rule.

A.If

f(x)=x

and

g(x)=\sin x

, then

(fg)&#039; = x\cos x + \sin x

B.The derivative of a constant times a function equals the constant times the derivative of the function.

(fg)&#039; = f&#039;g + fg&#039;

(fg)&#039; = f&#039;g&#039;

✅

💡 Difficulty: easy | ✅ Correct: D

📖 Explanation: All the listed statements follow from the product rule except the last one, which incorrectly claims that the derivative of a product equals the product of the derivatives; the correct rule includes an additional term.

Q3. If $f(a)=0$ for some $a$ and both $f$ and $g$ are differentiable at $a$ , what is $(f\cdot g)'(a)$ equal to?

f(a)g&#039;(a)

✅

g(a)f&#039;(a)

0

f&#039;(a)g&#039;(a)

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the product rule gives $(fg)' = f\,g' + g\,f'$ . Since $f(a)=0$ , the first term vanishes, leaving $(fg)'(a)=g(a)f'(a)$ .

Q4. Suppose $f'(x)=g(x)$ for all $x$ . Let $h(x)=f(x)g(x)$ . Which expression represents $h'(x)$ ?

f(x)g&#039;(x)+g(x)^2

✅

f&#039;(x)g(x)+f(x)g&#039;(x)

2g(x)^2

f(x)^2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using the product rule, $h' = f\,g' + g\,f'$ . Substituting $f' = g$ gives $h' = f\,g' + g\,g = f\,g' + g^2$ , which matches the first option.

Q5. Given the data at $x=a$ : $f(a)=2,\;g(a)=3,\;f'(a)=5,\;g'(a)=-1$ . Which of the following could represent $(f\cdot g)''(a)$ ?

2 f&#039;(a) g&#039;(a) + f(a) g&#039;&#039;(a) + g(a) f&#039;&#039;(a)

f&#039;(a) g&#039;(a) + f(a) g&#039;(a) + g(a) f&#039;(a)

f&#039;&#039;(a) g&#039;&#039;(a)

✅

f(a) g(a)

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Differentiating $(fg)' = f g' + g f'$ again yields $(fg)'' = f' g' + f g'' + g' f' + g f'' = 2 f' g' + f g'' + g f''$ . The first option correctly reflects this formula.

Q6. Find $dy/dx$ if $y = (3x^{2})(e^{x})$ .

6x e^{x} + 3x^{2} e^{x}

✅

3x^{2} e^{x}

6x e^{x}

3 e^{x}

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Applying the product rule: $(3x^{2})' = 6x$ and $(e^{x})' = e^{x}$ . Thus $y' = 6x e^{x} + 3x^{2} e^{x}$ , which is the first option.

Q7. For $y = (x+1)(x^{2}-3x+2)$ , which statement is true about using the product rule versus expanding first?

A.Both methods give the same derivative after simplification. ✅

B.Expanding first yields a different derivative.

C.The product rule is only valid for non‑polynomial functions.

D.Using the product rule avoids any algebraic errors.

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Whether you expand the product or apply the product rule, the resulting derivative is the same; the algebraic simplification may differ, but both approaches lead to the identical final expression.

Q8. In the proof of the product rule, which step is crucial when adding and subtracting $f(x+h)g(x)$ ?

A.Factoring

h

from the numerator.

B.Adding and subtracting the same term to split the limit. ✅

C.Applying continuity of

g

D.Using the chain rule.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The key technique is to add and subtract $f(x+h)g(x)$ so the numerator can be separated into two differences, each of which can be handled with the definition of derivative, allowing the limit to be split.

Q9. When differentiating $f\cdot g'$ again, which term arises from the product rule?

f&#039; g&#039;

✅

f g&#039;&#039;

g f&#039;&#039;

2 f&#039; g&#039;

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Differentiating $f\cdot g'$ gives $(f\cdot g')' = f' g' + f g''$ ; the term $f' g'$ is the one that directly results from applying the product rule to this product.

Q10. When is the product rule unnecessary because one factor is constant?

A.When both factors are variable.

B.When one factor equals zero.

C.When one factor is a non‑zero constant. ✅

D.When both factors are constants.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: If one factor is a constant, the derivative reduces to the constant multiplied by the derivative of the other factor, so the full product rule is not needed; the constant can be factored out immediately.

Q11. Apply the product rule to $y=(1+t)\sqrt{t}$ and simplify. Which expression is correct?

\dfrac{1+3t}{2\sqrt{t}}

✅

1+3t

\dfrac{1+t}{\sqrt{t}}

\dfrac{2+t}{\sqrt{t}}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using $(1+t)'=1$ and $(\sqrt{t})'=1/(2\sqrt{t})$ , we obtain $y' = (1+t)\frac{1}{2\sqrt{t}}+\sqrt{t} = \frac{1+3t}{2\sqrt{t}}$ , which matches the first option.

Q12. For $y = u(v)\cdot w(x)$ where $u$ depends on $v$ and $v$ depends on $x$ , which rules must be used to find $dy/dx$ ?

A.Only the product rule.

B.Only the chain rule.

C.Both the product and chain rules. ✅

D.No rule is needed.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The factor $u(v)$ requires the chain rule to differentiate with respect to $x$ , while the overall product with $w(x)$ requires the product rule; thus both rules are necessary.

Q13. Find $dy/dx$ for $y = \sin(x^{2})\,e^{3x}$ . Which expression is correct?

\cos(x^{2})\cdot 2x\cdot e^{3x} + \sin(x^{2})\cdot 3e^{3x}

✅

\sin(x^{2})\,e^{3x}

2x\cos(x^{2})\,e^{3x}

3e^{3x}\sin(x^{2})

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Differentiating $\sin(x^{2})$ gives $\cos(x^{2})\cdot 2x$ ; differentiating $e^{3x}$ gives $3e^{3x}$ . Applying the product rule yields the sum of these two terms, which is the first option.

Q14. For $y = x^{2}\ln x$ , at which $x$ does $y' = 0$ ?

x = 1

x = e^{-1/2}

✅

x = e^{1/2}

D.No solution in the domain.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Differentiating gives $y' = 2x\ln x + x$ . Setting this to zero yields $x(2\ln x + 1)=0$ ; since $x>0$ , we solve $2\ln x + 1 = 0$ giving $\ln x = -\tfrac12$ and $x = e^{-1/2}$ .

Q15. If $f(x)=e^{x^{2}}$ and $g(x)=\cos x$ , what is $(f\cdot g)'(\pi)$ ?

-2\pi e^{\pi^{2}}

✅

2\pi e^{\pi^{2}}

-e^{\pi^{2}}

e^{\pi^{2}}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using $(fg)' = f'g + fg'$ , we have $f'(\pi)=2\pi e^{\pi^{2}}$ , $g(\pi) = -1$ , and $g'(\pi)=0$ . Thus $(fg)'(\pi)=2\pi e^{\pi^{2}}(-1)+e^{\pi^{2}}(0) = -2\pi e^{\pi^{2}}$ .

Q16. Let $h(x) = (x^{3})(x^{3})$ . Using the product rule, what is $h'(x)$ and how does it compare to the power rule result for $x^{6}$ ?

6x^{5}

✅

3x^{5}

9x^{5}

x^{5}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Applying the product rule: $(x^{3})' = 3x^{2}$ ; thus $h' = 3x^{2}\cdot x^{3} + x^{3}\cdot 3x^{2} = 6x^{5}$ . The power rule for $x^{6}$ also yields $6x^{5}$ , confirming they agree.

Q17. If $f'(x_{0})=0$ and $g'(x_{0})=0$ , what is $(f\cdot g)'(x_{0})$ ?

f(x_{0})g&#039;(x_{0})

g(x_{0})f&#039;(x_{0})

0

✅

f&#039;(x_{0})g&#039;(x_{0})

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The product rule gives $(fg)' = f g' + g f'$ . Substituting the given zero derivatives, both terms vanish, leaving $(fg)'(x_{0}) = 0$ .

Q18. Which of the following functions requires the product rule rather than the quotient rule for differentiation?

\dfrac{x^{2}}{x+1}

(\sin x)(e^{x})

✅

\dfrac{e^{x}}{\sin x}

\dfrac{\ln x}{x}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The expression $(\sin x)(e^{x})$ is a direct product of two functions, so the product rule is the appropriate tool; the other choices are quotients and would use the quotient rule.

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