# Mathematics Questions and Answers – Logarithmic Differentiation

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This set of Mathematics 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) $$\frac{3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x log⁡2x}}{log⁡2x^{sin⁡3x}}$$

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⁡2xsin⁡3x $$\left (3 \,cos⁡3x \,log⁡(log⁡2x)+\frac{sin⁡3x}{x \,log⁡2x} \right )$$

2. Differentiate 4xex with respect to x.
a) xex e-x (x log⁡x+1)
b) -4xex-1 ex (x log⁡x+1)
c) 4xex ex (x log⁡x+1)
d) 4xex-1 ex (x log⁡x+1)

Explanation: Consider y=4xex
Applying log on both sides, we get
log⁡y=log⁡4xex
log⁡y=log⁡4+log⁡xex (∵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 9tan⁡3x with respect to x.
a) 9tan⁡3x (3 log⁡9 sec2⁡x)
b) 9tan⁡3x (3 log⁡3 sec2⁡⁡x)
c) 9tan⁡3x (3 log⁡9 sec⁡x)
d) -9tan⁡3x (3 log⁡9 sec2⁡⁡x)

Explanation: Consider y=9tan⁡3x
Applying log on both sides, we get
log⁡y=log⁡9tan⁡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 sec2⁡⁡x.log⁡9+0)
$$\frac{dy}{dx}$$=9tan⁡3x (3 log⁡9 sec2⁡x)
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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)

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(2e2x) with respect to x.
a) 14e2x x(2e2x) (2 log⁡x+$$\frac{1}{x}$$)
b) 14x(2e2x) (2 log⁡x+$$\frac{1}{x}$$)
c) 14e2x x(2e2x) (2 log⁡x-$$\frac{1}{x}$$)
d) 14e2x x(2e2x) (log⁡x-$$\frac{1}{x}$$)

Explanation: Consider y=7x(2e2x)
log⁡y=log⁡7x(2e2x)
log⁡y=log⁡7+log⁡x(2e2x)
log⁡y=log⁡7+2e2x log⁡x
Differentiating with respect to x on both sides, we get
$$\frac{1}{y} \frac{dy}{dx}$$=$$\frac{d}{dx}$$ (log⁡7+2e2x log⁡x)
$$\frac{1}{y} \frac{dy}{dx}$$=0+$$\frac{d}{dx}$$ (2e2x) log⁡x+$$\frac{d}{dx}$$ (log⁡x)2e2x (using u.v=u’ v+uv’)
$$\frac{1}{y} \frac{dy}{dx}$$=2e2x.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(2e2x) $$\left (4e^{2x} \,log⁡x+\frac{2e^{2x}}{x}\right)$$
$$\frac{dy}{dx}$$=14e2x x(2e2x) (2 log⁡x+$$\frac{1}{x}$$)
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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) $$x^{log⁡x^2 -1} (10x^4+log⁡2x^2)$$

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⁡xcot⁡x (1-log⁡(tan⁡x))
b) csc2⁡x.tan⁡xcot⁡x (1-log⁡(tan⁡x))
c) 2 csc2⁡x.tan⁡xcot⁡x (1+log⁡(tan⁡x))
d) 2tan⁡xcot⁡x (1-log⁡(tan⁡x))

Explanation: Consider y=2(tan⁡x)cot⁡x
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⁡xcot⁡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)

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) $$\frac{-2}{\sqrt{(3x-1)(x+1)}}$$

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 x3ex with respect to x.
a) 3e3x (3 log⁡x+$$\frac{1}{x}$$)
b) x3e3x.3e3x (3 log⁡x-$$\frac{1}{x}$$)
c) x3e3x (3 log⁡x+$$\frac{1}{x}$$)
d) x3e3x.3e3x (3 log⁡x+$$\frac{1}{x}$$)

Explanation: Consider y=x3e3x
Applying log on both sides, we get
log⁡y=3e3x log⁡x
Differentiating both sides with respect to x, we get
$$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} (3e^{3x}) log⁡x+\frac{d}{dx} (log⁡x)3e^{3x}$$
$$\frac{1}{y} \frac{dy}{dx}$$=3e3x.3.log⁡x+$$\frac{1}{x}$$ 3e3x
$$\frac{dy}{dx}$$=y(3e3x.3.log⁡x+$$\frac{1}{x}$$ 3e3x)
$$\frac{dy}{dx}$$=x3e3x.3e3x (3 log⁡x+$$\frac{1}{x}$$)

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