📝 Logarithmic differentiation technique (16 MCQs)

📖 From Calculus • 4. Topics in Differentiation • 16 questions available

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What is Logarithmic differentiation technique?

Definition:
Logarithmic differentiation is a powerful method used to differentiate functions of the form $y = f(x)^{g(x)}$ or products/quotients of many functions. By taking the natural logarithm of both sides, we use log properties to simplify exponents into multipliers and products into sums, making the differentiation process significantly easier before solving for $\frac{dy}{dx}$ .

Example:
Let $y = x^x$ . Take $\ln(y) = x \ln(x)$ . Differentiate implicitly: $\frac{1}{y}y' = 1 \cdot \ln(x) + x \cdot \frac{1}{x} = \ln(x) + 1$ . Thus, $y' = y(\ln(x) + 1) = x^x(\ln(x) + 1)$ .

Reason:
This technique bypasses the complexity of applying the product and chain rules repeatedly on complicated expressions. It transforms multiplicative relationships into additive ones, leveraging the simplicity of the derivative of $\ln(x)$ to handle variable exponents and large products.

Easy

Medium

Hard

📝 All Logarithmic differentiation technique MCQs

Q1. What is the derivative of $y = x^{r}$ using logarithmic differentiation?

rx^{r-1}

✅

rx^{r+1}

x^{r}

r

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Taking natural logs gives $\ln y = r\ln x$ . Differentiating both sides with respect to $x$ yields $\frac{1}{y}y' = \frac{r}{x}$ . Multiplying by $y = x^{r}$ results in $y' = rx^{r-1}$ , which corresponds to option A.

Q2. What is the derivative of $\ln|y|$ with respect to $x$ ?

y\frac{dy}{dx}

\frac{1}{y}\frac{dy}{dx}

✅

\frac{dy}{dx}

1\cdot\frac{dy}{dx}

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: By the chain rule, $\frac{d}{dx}\ln|y| = \frac{1}{y}\cdot\frac{dy}{dx}$ . This directly matches option B, confirming that the derivative of the natural log of a function is the function’s derivative divided by the function itself.

Q3. In the logarithmic derivative of $y = x^{2}\sqrt[3]{7x-14}(1+x^{2})^{-4}$ , which term corresponds to the factor $(1+x^{2})^{-4}$ ?

-\frac{8x}{1+x^{2}}

✅

\frac{8x}{1+x^{2}}

-\frac{4}{1+x^{2}}

\frac{4}{1+x^{2}}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The factor $(1+x^{2})^{-4}$ contributes $-4\ln(1+x^{2})$ to $\ln y$ . Differentiating gives $-4\cdot\frac{2x}{1+x^{2}} = -\frac{8x}{1+x^{2}}$ , which is option A.

Q4. For $f(x)=\dfrac{x^{3}}{\sqrt{x^{2}+1}}$ , which method generally requires fewer steps?

A.Product rule

B.Quotient rule

C.Logarithmic differentiation ✅

D.Both require the same number of steps

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Writing $f(x)=x^{3}(x^{2}+1)^{-1/2}$ allows taking logs, turning products into sums. Differentiating the log expression avoids repeated product‑rule applications, so logarithmic differentiation typically simplifies the process, making it the most efficient choice.

Q5. Using logarithmic differentiation, which expression correctly represents $f'(x)$ for $f(x)=(x^{2}+1)^{\sin x}$ ?

(x^{2}+1)^{\sin x}\!\left[\cos x\ln(x^{2}+1)-\frac{2x\sin x}{x^{2}+1}\right]

(x^{2}+1)^{\sin x}\!\left[\cos x\ln(x^{2}+1)+\frac{2x\sin x}{x^{2}+1}\right]

(x^{2}+1)^{\sin x}\!\left[\sin x\ln(x^{2}+1)+\frac{2x\cos x}{x^{2}+1}\right]

(x^{2}+1)^{\sin x}\!\left[\cos x\ln(x^{2}+1)+\frac{x\sin x}{x^{2}+1}\right]

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Taking logs gives $\ln f = \sin x\,\ln(x^{2}+1)$ . Differentiating yields $\frac{f'}{f}= \cos x\ln(x^{2}+1)+\sin x\cdot\frac{2x}{x^{2}+1}$ . Multiplying by $f$ gives the derivative shown in option D.

Q6. For $y = \dfrac{x-2}{\sqrt{x+3}}$ , for which $x$ values is logarithmic differentiation valid?

x>2

x>-3

✅

x\neq -3,2

D.All real numbers

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The original function requires the denominator $\sqrt{x+3}$ to be defined and non‑zero, so $x+3>0$ and $x\neq -3$ . This simplifies to $x>-3$ . Within this domain the logarithmic derivative exists, matching option B.

Q7. Using logarithmic differentiation, find $g'(x)$ for $g(x)=\dfrac{(x^{2}-1)^{5}}{(e^{x}\sqrt{x})^{3}}$ .

g(x)\!\left[\frac{10x}{x^{2}-1}-3-\frac{3}{2x}\right]

g(x)\!\left[\frac{10x}{x^{2}-1}+3+\frac{3}{2x}\right]

g(x)\!\left[\frac{5x}{x^{2}-1}-\frac{3}{x}\right]

✅

g(x)\!\left[\frac{5}{x^{2}-1}-\frac{3}{2x}\right]

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Write $\ln g =5\ln(x^{2}-1)-3[x+\tfrac12\ln x]$ . Differentiating gives $\frac{g'}{g}= \frac{10x}{x^{2}-1}-3-\frac{3}{2x}$ . Multiplying by $g$ yields the expression in option C.

Q8. In differentiating $h(x)=\left(\dfrac{\sin x}{x}\right)^{x^{2}}$ via logarithmic differentiation, which term captures the effect of the variable exponent?

x^{2}\frac{\cos x}{\sin x}

2x\ln\!\left(\frac{\sin x}{x}\right)

✅

\frac{\cos x}{\sin x}

\ln\!\left(\frac{\sin x}{x}\right)

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: After taking logs, $\ln h = x^{2}\ln\!\left(\frac{\sin x}{x}\right)$ . Differentiating, the derivative of the exponent $x^{2}$ contributes $2x\ln\!\left(\frac{\sin x}{x}\right)$ , which is the term described in option B.

Q9. What is $p'(x)$ for $p(x)=\ln\left|\dfrac{x^{3}-7x^{2}-3}{2+\sqrt{x}}\right|$ ?

\dfrac{3x^{2}-14x}{x^{3}-7x^{2}-3}-\dfrac{1}{2\sqrt{x}(2+\sqrt{x})}

✅

\dfrac{3x^{2}-14x}{x^{3}-7x^{2}-3}+\dfrac{1}{2\sqrt{x}(2+\sqrt{x})}

\dfrac{3x^{2}-14x}{x^{3}-7x^{2}-3}-\dfrac{1}{(2+\sqrt{x})}

\dfrac{3x^{2}-14x}{x^{3}-7x^{2}-3}+\dfrac{1}{(2+\sqrt{x})}

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Using $\frac{d}{dx}\ln|u| = u'/u$ , the numerator derivative is $\frac{3x^{2}-14x}{x^{3}-7x^{2}-3}$ . The denominator contributes $-\frac{1}{2\sqrt{x}(2+\sqrt{x})}$ . Combining gives the expression in option A.

Q10. Which function cannot be differentiated directly using logarithmic differentiation?

y=e^{x^{2}}

y=\ln(x^{3}+1)

✅

y=x^{\sin x}

y=\sin(x)^{\cos x}

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Logarithmic differentiation is most useful for functions expressed as a variable base raised to a variable exponent. The function $\ln(x^{3}+1)$ is already a logarithm; applying another log would complicate rather than simplify the process, making option B unsuitable.

Q11. What is the derivative of $y = \ln|x|$ ?

\frac{1}{x}

✅

-\frac{1}{x}

0

1

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Differentiating $\ln|x|$ gives $\frac{d}{dx}\ln|x| = \frac{1}{x}$ for all $x\neq0$ . This matches option A.

Q12. Using logarithmic differentiation, what is $\frac{dy}{dx}$ for $y=\sqrt{x^{2}+1}$ ?

\frac{x}{\sqrt{x^{2}+1}}

✅

\frac{2x}{\sqrt{x^{2}+1}}

\frac{x}{2\sqrt{x^{2}+1}}

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

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Write $\ln y = \tfrac12\ln(x^{2}+1)$ . Differentiating gives $\frac{y'}{y}= \frac{x}{x^{2}+1}$ . Multiplying by $y = \sqrt{x^{2}+1}$ yields $y' = \frac{x}{\sqrt{x^{2}+1}}$ , which is option A.

Q13. For $f(x)=\exp\!\big(x\ln(x^{2}+3)\big)$ , which statement about its derivative is true?

f&#039;(x)=f(x)\big[\ln(x^{2}+3)+\frac{2x^{2}}{x^{2}+3}\big]

✅

f&#039;(x)=f(x)\big[\ln(x^{2}+3)+\frac{2x}{x^{2}+3}\big]

f&#039;(x)=f(x)\big[\frac{1}{x}+\ln(x^{2}+3)\big]

f&#039;(x)=f(x)\frac{2x}{x^{2}+3}

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Since $f = (x^{2}+3)^{x}$ , $\ln f = x\ln(x^{2}+3)$ . Differentiating gives $\frac{f'}{f}= \ln(x^{2}+3)+x\frac{2x}{x^{2}+3}= \ln(x^{2}+3)+\frac{2x^{2}}{x^{2}+3}$ . Multiplying by $f$ yields the expression in option A.

Q14. Using logarithmic differentiation, find $k'(x)$ for $k(x)=\left(\frac{1+x}{1-x}\right)^{5}$ .

k(x)\frac{10}{1-x^{2}}

k(x)\frac{5}{1-x^{2}}

k(x)\frac{10x}{1-x^{2}}

k(x)\frac{5x}{1-x^{2}}

✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: Taking logs: $\ln k =5[\ln(1+x)-\ln(1-x)]$ . Differentiating gives $\frac{k'}{k}=5\big[\frac{1}{1+x}+\frac{1}{1-x}\big]=\frac{10}{1-x^{2}}$ . Hence $k'=k\frac{10}{1-x^{2}}$ , which is option D.

Q15. If $z(x)=\ln\!\big(\frac{x^{x}}{e^{x}}\big)$ , what simplifies to $z'(x)$ ?

\ln x

✅

\ln x-1

\frac{1}{x}

\frac{1}{x}-1

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Rewrite $z = x\ln x - x$ . Differentiating yields $z' = \ln x + 1 - 1 = \ln x$ . Thus the derivative simplifies to $\ln x$ , which is option A.

Q16. When applying logarithmic differentiation to $y=(\sin x)^{\cos x}$ , which step is essential?

A.Taking the natural log of both sides ✅

B.Differentiating the numerator only

C.Using the quotient rule

D.Applying the product rule to the exponent only

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The method starts by writing $\ln y = \cos x\,\ln(\sin x)$ . This conversion to a log form is crucial because it turns the variable base and exponent into a product, allowing straightforward differentiation. Hence, taking the natural log is the essential first step.

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