📝 Derivative of ln x and log base b (16 MCQs)

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

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What is Derivative of ln x and log base b?

Definition:
The derivative of the natural logarithm function $\ln(x)$ is uniquely simple, given by $\frac{1}{x}$ for $x > 0$ . For a logarithm with base $b$ , denoted $\log_b(x)$ , the derivative is $\frac{1}{x \ln(b)}$ . This relationship arises because $\log_b(x)$ can be rewritten as $\frac{\ln(x)}{\ln(b)}$ , making $\frac{1}{\ln(b)}$ a constant multiplier in the differentiation process.

Example:
Differentiate $y = \ln(5x)$ . Using the chain rule, $y' = \frac{1}{5x} \cdot 5 = \frac{1}{x}$ . Alternatively, for $y = \log_{10}(x)$ , the derivative is $y' = \frac{1}{x \ln(10)}$ , which is approximately $\frac{0.434}{x}$ .

Reason:
Knowing these specific forms allows for quick calculation of rates of change in scientific formulas. The simplicity of $\frac{1}{x}$ for natural logs makes them preferred in calculus, while the base conversion factor handles other logarithmic bases efficiently.

Easy

Medium

Hard

📝 All Derivative of ln x and log base b MCQs

Q1. What is the derivative of \ $\\ln x\$ for \ $x>0\$ ?

A.\

1/x\

✅

B.\

x\

C.\

\\ln x\

D.\

0\

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: The derivative of the natural logarithm follows directly from the definition of the logarithmic function. Differentiating \ $\\ln x\$ yields \ $\\frac{d}{dx}\\ln x = \\frac{1}{x}\$ . This result holds for every positive \ $x\$ , confirming option A as the correct choice.

Q2. What is the derivative of \ $\\log_{b} x\$ with respect to \ $x\$ (for \ $b>0, b\\neq1\$ )?

A.\

b^{x}\

B.\

\\frac{1}{xb}\

C.\

\\frac{1}{x\\ln b}\

✅

D.\

\\frac{\\ln b}{x}\

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Using the change‑of‑base formula \ $\\log_{b}x=\\frac{\\ln x}{\\ln b}\$ and differentiating, we obtain \ $\\frac{d}{dx}\\log_{b}x = \\frac{1}{\\ln b}\\cdot\\frac{1}{x}=\\frac{1}{x\\ln b}\$ . This matches option C, while the other choices do not reflect the correct application of the chain rule.

Q3. Does the graph of \ $y=\\ln x\$ contain any horizontal tangent lines?

A.Yes at \

x=1\

B.No ✅

C.Yes at \

x=e\

D.Yes at \

x=0\

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: A horizontal tangent would require a slope of zero. The derivative of \ $\\ln x\$ is \ $1/x\$ , which is never zero for any real \ $x>0\$ . Consequently, the function has no horizontal tangents, making option B the correct inference.

Q4. What is the sign of the slope of \ $y=\\ln|x|\$ at \ $x=-3\$ ?

A.Positive

B.Zero

C.Undefined

D.Negative ✅

💡 Difficulty: medium | ✅ Correct: D

📖 Explanation: The derivative of \ $\\ln|x|\$ is \ $1/x\$ . Substituting \ $x=-3\$ gives \ $1/(-3)=-\\frac{1}{3}\$ , a negative value. Hence the slope is negative, confirming option D as the correct answer.

Q5. If \ $f(x)=\\ln(x^{2})\$ , what is \ $f'(1)\$ ?

A.2 ✅

B.1

C.0

D.-2

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: First rewrite \ $\\ln(x^{2})=2\\ln x\$ . Differentiating gives \ $f'(x)=2\\cdot\\frac{1}{x}=\\frac{2}{x}\$ . Evaluating at \ $x=1\$ yields \ $f'(1)=2\$ . Therefore option A correctly represents the derivative at the specified point.

Q6. Let \ $h(x)=\\ln|x|+\\ln|x-1|\$ . Which value of \ $x\$ makes \ $h'(x)=0\$ ?

A.\

0\

B.\

\\tfrac12\

✅

C.\

1\

D.None

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Differentiating gives \ $h'(x)=\\frac{1}{x}+\\frac{1}{x-1}=\\frac{2x-1}{x(x-1)}\$ . Setting the numerator to zero yields \ $2x-1=0\$ → \ $x=\\tfrac12\$ . This point is within the domain (excluding 0 and 1), so option B is correct.

Q7. At \ $x=1\$ , which derivative is larger: \ $\\frac{d}{dx}\\ln(x^{2}+1)\$ or \ $\\frac{d}{dx}\\ln(x^{2})\$ ?

A.Derivative of \

\\ln(x^{2}+1)\

is larger

B.Derivative of \

\\ln(x^{2})\

is larger ✅

C.They are equal

D.Cannot compare

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Compute each derivative: \ $\\frac{d}{dx}\\ln(x^{2}+1)=\\frac{2x}{x^{2}+1}=1\$ at \ $x=1\$ . For \ $\\ln(x^{2})\$ , the derivative is \ $\\frac{2}{x}=2\$ at \ $x=1\$ . Since \ $2>1\$ , the derivative of \ $\\ln(x^{2})\$ is larger, confirming option B.

Q8. Evaluate \ $\\displaystyle\\lim_{x\\to0}\\frac{\\ln(1+x)}{x}\$ using differentiation.

A.1 ✅

B.0

C.Infinity

D.Does not exist

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Applying L'Hôpital's Rule, differentiate numerator and denominator: \ $\\frac{d}{dx}\\ln(1+x)=\\frac{1}{1+x}\$ and \ $\\frac{d}{dx}x=1\$ . The limit becomes \ $\\lim_{x\\to0}\\frac{1}{1+x}=1\$ . Thus the limit equals 1, corresponding to option A.

Q9. For \ $f(x)=\\frac{\\ln x}{x}\$ , at which \ $x\$ does \ $f'(x)=0\$ ?

A.\

x=1\

B.\

x=e\

✅

C.\

x=\\sqrt e\

D.No solution

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Differentiating gives \ $f'(x)=\\frac{1-\\ln x}{x^{2}}\$ . Setting the numerator to zero yields \ $\\ln x=1\$ → \ $x=e\$ . Therefore the derivative vanishes at \ $x=e\$ , making option B correct.

Q10. Simplify the derivative of \ $p(x)=\\log_{2}(x^{3})-\\ln x\$ .

A.\

\\frac{3}{x\\ln2}-\\frac{1}{x}\

✅

B.\

\\frac{3}{x}-\\frac{1}{x\\ln2}\

C.\

\\frac{3\\ln2-1}{x}\

D.\

\\frac{3-\\ln2}{x}\

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Using the change‑of‑base formula, \ $\\log_{2}(x^{3})=\\frac{\\ln(x^{3})}{\\ln2}=\\frac{3\\ln x}{\\ln2}\$ . Differentiating gives \ $\\frac{3}{x\\ln2}\$ . Subtracting the derivative of \ $\\ln x\$ (which is \ $1/x\$ ) yields \ $\\frac{3}{x\\ln2}-\\frac{1}{x}\$ , matching option A.

Q11. Why is the derivative of \ $\\ln|x|\$ equal to \ $1/x\$ for \ $x<0\$ ?

A.Because the derivative of \

-x\

is \

-1\

B.Because the chain rule gives \

\\frac{1}{-x}\\cdot(-1)=\\frac{1}{x}\

✅

C.Because the absolute value removes the sign

D.It is not defined for \

x<0\

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: When \ $x<0\$ , \ $|x|=-x\$ . Applying the chain rule to \ $\\ln(-x)\$ yields \ $\\frac{1}{-x}\\cdot\\frac{d}{dx}(-x)=\\frac{1}{-x}\\cdot(-1)=\\frac{1}{x}\$ . Hence the derivative retains the same form \ $1/x\$ despite the negative argument, confirming option B.

Q12. Apply the chain rule to \ $y=\\ln\\big((\\sin x)^{2}+e^{x}\\big)\$ . What is \ $dy/dx\$ ?

A.\

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

✅

B.\

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

C.\

\\frac{e^{x}}{(\\sin x)^{2}+e^{x}}\

D.\

\\frac{2\\sin x\\cos x-e^{x}}{(\\sin x)^{2}+e^{x}}\

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: Let \ $u=(\\sin x)^{2}+e^{x}\$ . Then \ $y=\\ln u\$ and \ $dy/dx=\\frac{1}{u}\\cdot du/dx\$ . Differentiating \ $u\$ gives \ $du/dx=2\\sin x\\cos x+e^{x}\$ . Substituting back yields \ $dy/dx=\\frac{2\\sin x\\cos x+e^{x}}{(\\sin x)^{2}+e^{x}}\$ , which is option A.

Q13. If \ $y=\\ln(u)\$ where \ $u>0\$ and differentiable, express \ $dy/dx\$ in terms of \ $u\$ and \ $du/dx\$ .

A.\

u\\cdot\\frac{du}{dx}\

B.\

\\frac{du}{dx}\

C.\

\\frac{1}{u}\\cdot\\frac{du}{dx}\

✅

D.\

\\ln\\big(\\frac{du}{dx}\\big)\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Applying the chain rule to the composition \ $\\ln(u(x))\$ gives \ $\\frac{dy}{dx}=\\frac{1}{u}\\cdot\\frac{du}{dx}\$ . This formula captures the rate of change of the outer logarithmic function multiplied by the inner derivative, matching option C.

Q14. Determine the sign of the second derivative of \ $f(x)=\\ln|x|\\cdot\\ln|x-2|\$ at \ $x=3\$ .

A.Positive

B.Negative ✅

C.Zero

D.Undefined

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: First derivative: \ $f'(x)=\\frac{\\ln|x-2|}{x}+\\frac{\\ln|x|}{x-2}\$ . Differentiating again and evaluating at \ $x=3\$ yields a negative value because both logarithmic terms are positive while the dominant term \ $-\\frac{\\ln|x|}{(x-2)^{2}}\$ is negative and larger in magnitude. Hence the second derivative is negative, confirming option B.

Q15. Using the change‑of‑base formula, which step correctly derives \ $\\frac{d}{dx}\\log_{b}x\$ ?

A.Apply definition of logarithm

B.Rewrite as \

\\frac{\\ln x}{\\ln b}\

then differentiate ✅

C.Use product rule

D.No proof needed

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Starting from \ $\\log_{b}x=\\frac{\\ln x}{\\ln b}\$ and noting that \ $\\ln b\$ is constant, differentiating yields \ $\\frac{1}{\\ln b}\\cdot\\frac{1}{x}=\\frac{1}{x\\ln b}\$ . This logical sequence matches option B and establishes the derivative formula.

Q16. Find the equation of the tangent line to \ $y=\\ln x\$ at \ $x=4\$ and state its y‑intercept.

A.\

y=\\frac{1}{4}x-\\ln4\

B.\

y=\\frac{1}{4}x+\\ln4-1\

C.\

y=\\frac{1}{4}x+(\\ln4-1)\

✅

D.\

y=\\frac{1}{4}x-(\\ln4-1)\

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The slope at \ $x=4\$ is \ $m=1/4\$ . Using point‑slope form with point \ $(4,\\ln4)\$ : \ $y-\\ln4=\\frac{1}{4}(x-4)\$ → \ $y=\\frac{1}{4}x+\\ln4-1\$ . The y‑intercept occurs at \ $x=0\$ giving \ $y=\\ln4-1\$ . This corresponds to option C.

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