Logarithmic Differentiation - Class 12 Maths MCQ

This set of Class 12 Maths Chapter 5 Multiple Choice Questions & Answers (MCQs) focuses on “Logarithmic Differentiation”.

1. Differentiate (log⁡2x)^sin⁡3x with respect to x.
a) (3 cos⁡3x log⁡(log⁡2x)+\(\frac{sin⁡3x}{x log⁡2x}\))
b) \(log⁡2x^{sin⁡3x} \,(3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x})\)
c) –\((3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x log⁡2x})\)
d) \(\frac{3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x log⁡2x}}{log⁡2x^{sin⁡3x}}\)
View Answer

Answer: b
Explanation: Consider y=\((log⁡2x)^{sin⁡3x}\)
Applying log on both sides, we get
log⁡y=\(log⁡(log⁡2x)^{sin⁡3x}\)
log⁡y=sin⁡3x log⁡(log⁡2x)
Differentiating with respect to x, we get
\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x)\frac{d}{dx} (sin⁡3x)+sin⁡3x \frac{d}{dx} \,(log⁡(log⁡2x))\)
By using chain rule, we get
\(\frac{1}{y} \,\frac{dy}{dx}=log⁡(log⁡2x).3 \,cos⁡3x+sin⁡3x.\frac{1}{log⁡2x}.\frac{1}{2x}.2 \,(∵u.v=u’ \,v+uv’)\)
\(\frac{dy}{dx}\)=y(3 cos⁡3x log⁡(log⁡2x)+\(\frac{sin⁡3x}{x \,log⁡2x}\))
∴\(\frac{dy}{dx}\)=log⁡2x^sin⁡3x \(\left (3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x} \right )\)

2. Differentiate 4x^{e^x} with respect to x.
a) x^{e^x} e^-x (x log⁡x+1)
b) -4x^{e^x-1} e^x (x log⁡x+1)
c) 4x^{e^x} e^x (x log⁡x+1)
d) 4x^{e^x-1} e^x (x log⁡x+1)
View Answer

Answer: d
Explanation: Consider y=4x^{e^x}
Applying log on both sides, we get
log⁡y=log⁡4x^{e^x}
log⁡y=log⁡4+log⁡x^{e^x} (∵log⁡ab=log⁡a+log⁡b)
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}=0+\frac{d}{dx}(e^x \,log⁡x)(∵log⁡a^b=b \,log⁡a)\)
\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \,(e^x) \,log⁡x+e^{x} \,\frac{d}{dx} \,(log⁡x)\)
\(\frac{dy}{dx}=y(e^x log⁡x+\frac{e^x}{x})\)
\(\frac{dy}{dx}=\frac{4x^{e^{x}}e^x \,(x log⁡x+1)}{x}=4x^{e^x-1} \,e^x \,(x log⁡x+1)\).

3. Differentiate 9^tan⁡3x with respect to x.
a) 9^tan⁡3x (3 log⁡9 sec^2⁡x)
b) 9^tan⁡3x (3 log⁡3 sec^2⁡⁡x)
c) 9^tan⁡3x (3 log⁡9 sec⁡x)
d) -9^tan⁡3x (3 log⁡9 sec^2⁡⁡x)
View Answer

Answer: a
Explanation: Consider y=9^tan⁡3x
Applying log on both sides, we get
log⁡y=log⁡9^tan⁡3x
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \)(tan⁡3x.log⁡9)
\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} \,(tan⁡3x) \,log⁡9+\frac{d}{dx} \,(log⁡9).tan3x \,(∵ Using \,u.v=u’ \,v+uv’)\)
\(\frac{dy}{dx}\)=y(3 sec^2⁡⁡x.log⁡9+0)
\(\frac{dy}{dx}\)=9^tan⁡3x (3 log⁡9 sec^2⁡x)

4. Differentiate (cos⁡3x)^3x with respect to x.
a) (cos⁡3x)^x (3 log⁡(cos⁡3x) – 9x tan⁡3x)
b) (cos⁡3x)^3x (3 log⁡(cos⁡3x) + 9x tan⁡3x)
c) (cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)
d) (cos⁡3x)^3x (log⁡(cos⁡3x) + 9 tan⁡3x)
View Answer

Answer: c
Explanation: Consider y=(cos⁡3x)^3x
Applying log on both sides, we get
log⁡y=log⁡(cos⁡3x)^3x
log⁡y=3x log⁡(cos⁡3x)
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}=\frac{d}{dx} (3x \,log⁡(cos⁡3x))\)
By using u.v=u’ v+uv’, we get
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{d}{dx} \,(3x) \,log⁡(cos⁡3x)+\frac{d}{dx} \,(log⁡(cos⁡3x)).3x\)
\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \frac{d}{dx} \,(cos⁡3x).3x)\)
\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).\frac{d}{dx}(3x).3x)\)
\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) + \(\frac{1}{cos⁡3x} \,. \,(-sin⁡3x).3.3x)\)
\(\frac{dy}{dx}\)=y(3 log⁡(cos⁡3x) – 9x tan⁡3x)
\(\frac{dy}{dx}\)=(cos⁡3x)^3x (3 log⁡(cos⁡3x) – 9x tan⁡3x)

5. Differentiate 7x^{(2e^2x)} with respect to x.
a) 14e^2x x^{(2e^2x)} (2 log⁡x+\(\frac{1}{x}\))
b) 14x^{(2e^2x)} (2 log⁡x+\(\frac{1}{x}\))
c) 14e^2x x^{(2e^2x)} (2 log⁡x-\(\frac{1}{x}\))
d) 14e^2x x^{(2e^2x)} (log⁡x-\(\frac{1}{x}\))
View Answer

Answer: a
Explanation: Consider y=7x^{(2e^2x)}
log⁡y=log⁡7x^{(2e^2x)}
log⁡y=log⁡7+log⁡x^{(2e^2x)}
log⁡y=log⁡7+2e^2x log⁡x
Differentiating with respect to x on both sides, we get
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{d}{dx}\) (log⁡7+2e^2x log⁡x)
\(\frac{1}{y} \frac{dy}{dx}\)=0+\(\frac{d}{dx}\) (2e^2x) log⁡x+\(\frac{d}{dx}\) (log⁡x)2e^2x (using u.v=u’ v+uv’)
\(\frac{1}{y} \frac{dy}{dx}\)=2e^2x.2.log⁡x+\(\frac{2e^{2x}}{x}\)
\(\frac{dy}{dx}\)=y\( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)
\(\frac{dy}{dx}\)=7x^{(2e^2x)} \( \left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)\)
\(\frac{dy}{dx}\)=14e^2x x^{(2e^2x)} (2 log⁡x+\(\frac{1}{x}\))

Note: Join free Sanfoundry classes at Telegram or Youtube

6. Differentiate \(e^{4x^5}.2x^{log⁡x^2}\) with respect to x.
a) \(e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)
b) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5+log⁡2x^2)\)
c) \(4e^{4x^5}.x^{log⁡x^2-1} (10x^5-log⁡2x^2)\)
d) \(x^{log⁡x^2 -1} (10x^4+log⁡2x^2)\)
View Answer

Answer: b
Explanation: Consider y=\(e^{4x^5}+2x^{log⁡x^2}\)
Applying log on both sides, we get
log⁡y=\(log⁡e^{4x^5} \,+ \,log⁡2x^{log⁡x^2}\)
log⁡y=\(4x^5+log⁡x^2 \,. \,log⁡2x\)
log⁡y=\(4x^5+2 \,log⁡x \,log⁡2x\)
Differentiating with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+2(\frac{d}{dx} \,(log⁡x) \,log⁡2x+\frac{d}{dx} \,(log⁡2x) \,log⁡x)\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+2\left (\frac{log⁡2x}{x}+\frac{1}{2x}.2.log⁡x\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x+log⁡x)}{x}\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(20x^4+\frac{2(log⁡2x^2)}{x}\)
\(\frac{dy}{dx}\)=\(y(20x^4+\frac{2(log⁡2x^2)}{x})\)
\(\frac{dy}{dx}\)=\(e^{4x^5}.2x^{log⁡x^2} (20x^4+\frac{2(log⁡2x^2)}{x})\)
\(\frac{dy}{dx}\)=\(4e^{4x^5}.x^{log⁡x^2 -1} (10x^5+log⁡2x^2)\)

7. Differentiate 2(tan⁡x)^cot⁡x with respect to x.
a) 2 csc²⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))
b) csc²⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))
c) 2 csc²⁡x.tan⁡x^cot⁡x (1+log⁡(tan⁡x))
d) 2tan⁡x^cot⁡x (1-log⁡(tan⁡x))
View Answer

Answer: a
Explanation: Consider y=2(tan⁡x)^cot⁡x
Applying log in both sides,
log⁡y=log⁡2(tan⁡x)^cot⁡x
log⁡y=log⁡2+log⁡(tan⁡x)^cot⁡x
log⁡y=log⁡2+cot⁡x log⁡(tan⁡x)
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}=0+\frac{d}{dx} \,(cot⁡x) \,log⁡(tan⁡x)+cot⁡x \frac{d}{dx} \,(log⁡(tan⁡x))\)
\(\frac{1}{y} \frac{dy}{dx}=-csc^{2⁡}x.log⁡(tan⁡x)+cot⁡x.\frac{1}{tan⁡x}.sec^{2⁡}x\)
\(\frac{dy}{dx} = y\left(-csc^{2⁡x}.log⁡(tan⁡x)+\frac{(1+tan^{2⁡x})}{tan^{2⁡x}}\right)\)
\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x \(\left (-csc^{2⁡x} log⁡(tan⁡x)+cot^{2⁡x}+1 \right )\)
\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x \((-csc^{2⁡x} log⁡(tan⁡x)+csc^{2⁡x})\)
\(\frac{dy}{dx}\)=2(tan⁡x)^cot⁡x (csc²⁡x (1-log⁡(tan⁡x))
∴\(\frac{dy}{dx}\)=2 csc²⁡x.tan⁡x^cot⁡x (1-log⁡(tan⁡x))

8. Differentiate (3 cos⁡x)^x with respect to x.
a) (3 cos⁡x)^x (log⁡(3 cos⁡x)+x tan⁡x)
b) (3 cos⁡x)^x (log⁡(3 cos⁡x)+tan⁡x)
c) (cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)
d) (3 cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)
View Answer

Answer: d
Explanation: Consider y=(3 cos⁡x)^x
Applying log on both sides, we get
log⁡y=log⁡(3 cos⁡x)^x
log⁡y=x log⁡(3 cos⁡x)
log⁡y=x(log⁡3+log⁡(cos⁡x))
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx} =\frac{d}{dx} (x log⁡3)+\frac{d}{dx} (x) log⁡(cos⁡x)+\frac{d}{dx}(log⁡(cos⁡x)).x\)
\(\frac{1}{y} \frac{dy}{dx}=log⁡3+log⁡(cos⁡x)+\frac{1}{cos⁡x}.-sin⁡x.x\)
\(\frac{1}{y} \frac{dy}{dx}\)=log⁡3+log⁡(cos⁡x)-x tan⁡x
\(\frac{dy}{dx}\)=y(log⁡(3 cos⁡x)-x tan⁡x)
\(\frac{dy}{dx}\)=(3 cos⁡x)^x (log⁡(3 cos⁡x)-x tan⁡x)

9. Differentiate \(\sqrt{\frac{x+1}{3x-1}}\) with respect to x.
a) \(\frac{-2}{(3x-1)\sqrt{(3x-1)(x+1)}}\)
b) \(\frac{2}{(3x-1)\sqrt{(3x-1)(x+1)}}\)
c) \(\frac{1}{(3x-1)\sqrt{(3x-1)(x+1)}}\)
d) \(\frac{-2}{\sqrt{(3x-1)(x+1)}}\)
View Answer

Answer: a
Explanation: Consider y=\(\sqrt{\frac{x+1}{3x-1}}\)
Applying log to both sides, we get
log⁡y=log⁡\(\sqrt{\frac{x+1}{3x-1}}\)
log⁡y=\(\frac{1}{2} log⁡\left (\frac{x+1}{3x-1}\right )\)
log⁡y=\(\frac{1}{2}\) (log⁡(x+1)-log⁡(3x-1))
Differentiating with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{d}{dx} (log⁡(x+1))-\frac{d}{dx} (log⁡(3x-1))\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{1}{x+1}-\frac{3}{3x-1}\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{3x-1-3x-3}{(x+1)(3x-1)})\right )\)
\(\frac{1}{y} \frac{dy}{dx}\)=\(\frac{1}{2}\left (\frac{-4}{(x+1)(3x-1)}\right )\)
\(\frac{dy}{dx}\)=\(\sqrt{\frac{x+1}{3x-1}} \left (\frac{-2}{(x+1)(3x-1)}\right )\)
\(\frac{dy}{dx}\)=\(\frac{-2}{(3x-1) \sqrt{(3x-1)(x+1)}}\)

10. Differentiate x^{3e^x} with respect to x.
a) 3e^3x (3 log⁡x+\(\frac{1}{x}\))
b) x^{3e^3x}.3e^3x (3 log⁡x-\(\frac{1}{x}\))
c) x^{3e^3x} (3 log⁡x+\(\frac{1}{x}\))
d) x^{3e^3x}.3e^3x (3 log⁡x+\(\frac{1}{x}\))
View Answer

Answer: d
Explanation: Consider y=x^{3e^3x}
Applying log on both sides, we get
log⁡y=3e^3x log⁡x
Differentiating both sides with respect to x, we get
\(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3e^{3x}) log⁡x+\frac{d}{dx} (log⁡x)3e^{3x}\)
\(\frac{1}{y} \frac{dy}{dx}\)=3e^3x.3.log⁡x+\(\frac{1}{x}\) 3e^3x
\(\frac{dy}{dx}\)=y(3e^3x.3.log⁡x+\(\frac{1}{x}\) 3e^3x)
\(\frac{dy}{dx}\)=x^{3e^3x}.3e^3x (3 log⁡x+\(\frac{1}{x}\))

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

« Prev - Class 12 Maths MCQ – Exponential and Logarithmic Functions

» Next - Class 12 Maths MCQ – Derivatives of Functions in Parametric Forms

Related Posts:

Recommended Articles: