Mathematics Questions and Answers – Exponential and Logarithmic Functions

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Exponential and Logarithmic Functions”.

1. Differentiate 8e-x+2ex w.r.t x.
a) 2e-x+8ex
b) 2ex+8e-x
c) 2e-x-8ex
d) 2ex-8e-x
View Answer

Answer: d
Explanation: To solve:y=(8e-x+2ex)
Differentiating w.r.t x we get,
\(\frac{dy}{dx}\)=8(-e-x+2ex)
∴\(\frac{dy}{dx}\)=2ex-8e-x.
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2. Differentiate 8ecos2x w.r.t x.
a) 16 sin⁡2x ecos2x
b) -16 sin⁡2x ecos2x
c) -16 sin⁡2x e-cos⁡2x
d) 16 sin⁡2x e-cos⁡2x
View Answer

Answer: b
Explanation: Consider y=8ecos2x
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}=\frac{d}{dx}\)(8ecos2x)
=8ecos2x\(\frac{d}{dx}\) (cos⁡2x)
=8ecos2x(-sin⁡2x)\(\frac{d}{dx}\) (2x)
=8ecos2x(-sin⁡2x)(2)
∴\(\frac{dy}{dx}\)=-16 sin⁡2x ecos2x

3. Differentiate 3 sin-1⁡(e2x) w.r.t x.
a) \(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)
b) \(\frac{2e^2x}{\sqrt{1-e^{4x}}}\)
c) –\(\frac{6e^2x}{\sqrt{1-e^{4x}}}\)
d) \(\frac{6e^{-2x}}{\sqrt{1-e^{4x}}}\)
View Answer

Answer: a
Explanation: Consider y=3 sin-1⁡(e2x)
\(\frac{dy}{dx}=\frac{d}{dx}\)(3 sin-1⁡(e2x))
\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\frac{d}{dx}\)(e2x)
\(\frac{dy}{dx}=\left (\frac{3}{\sqrt{1-(e^{2x})^2}}\right )\)2e2x
∴\(\frac{dy}{dx}\)=\(\frac{6e^{2x}}{\sqrt{1-e^{4x}}}\)
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4. Differentiate log⁡(log⁡x5) w.r.t x.
a) –\(\frac{5}{x log⁡x^5}\)
b) \(\frac{1}{log⁡x^5}\)
c) \(\frac{5}{x log⁡x^5}\)
d) –\(\frac{1}{x log⁡x^5}\)
View Answer

Answer: c
Explanation: Consider y=(log⁡(log⁡(x5)))
\(\frac{dy}{dx}=\frac{1}{log⁡x^5} \frac{d}{dx} (log⁡x^5)\)
\(\frac{dy}{dx}=\frac{1}{log⁡x^5}.\frac{1}{x^5}.\frac{d}{dx} (x^5)\)
\(\frac{dy}{dx}=\frac{1}{log⁡x^5}.\frac{1}{x^5}.5x^4\)
∴\(\frac{dy}{dx}=\frac{5}{x log⁡x^5}\)

5. Differentiate 3e3x3 w.r.t x.
a) 27x-2 e3x3
b) 27x2 e3x3
c) -27x2 e3x3
d) -27x-2 e3x3
View Answer

Answer: b
Explanation: Consider y=3e3x3
\(\frac{dy}{dx}\)=\(\frac{d}{dx}\)(3e3x3)
\(\frac{dy}{dx}\)=3e3x3 \(\frac{d}{dx}\)(3x3)
\(\frac{dy}{dx}\)=3e3x3 (3(3x2))
\(\frac{dy}{dx}\)=27x2 e3x3.
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6. Differentiate 5ex2 tan⁡x w.r.t x.
a) 5ex2 (1+tan⁡x)2
b) -5ex2 (1+tan⁡x)2
c) 5ex2 (1-tan⁡x)2
d) -5ex2 (1-tan⁡x)2
View Answer

Answer: a
Explanation: Consider y=5ex2 tan⁡x
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=tan⁡x \(\frac{d}{dx}\) (5ex2)+5ex2 \(\frac{d}{dx}\) (tan⁡x)
\(\frac{dy}{dx}\)=tan⁡x (5ex2.2x)+5ex2 (sec2⁡x)
\(\frac{dy}{dx}\)=5ex2 (2x tan⁡x+sec2⁡x)
\(\frac{dy}{dx}\)=5ex2 (1+tan2⁡x+2x tan⁡x)
\(\frac{dy}{dx}\)=5ex2 (1+tan⁡x)2

7. Differentiate log⁡(e5x3) w.r.t x.
a) \(\frac{-15x^2}{e^{5x^3}}\)
b) \(\frac{15x^2}{e^{5x^3}}\)
c) 15x2
d) -15x2
View Answer

Answer: c
Explanation: Consider y=log⁡(e5x3)
y=5x3 (∴log⁡ex=x)
⇒\(\frac{dy}{dx}\)=\(\frac{d}{dx} (5x^3)\)
∴\(\frac{dy}{dx}=5(3x^2)=15x^2\)
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8. Differentiate 7 log⁡(x4.5ex3) w.r.t x.
a) \(\frac{7(4+3x^3)}{x^2}\)
b) \(\frac{7(4-3x^3)}{x}\)
c) –\(\frac{7(4+3x^3)}{x}\)
d) \(\frac{7(4+3x^3)}{x}\)
View Answer

Answer: d
Explanation: Consider y=7 log⁡(x4.5ex3)
y=\(7(log⁡x^4 +log⁡5e^{x^3})\)
y=\(7(4 log⁡x+log⁡5e^{x^3})\)
\(\frac{dy}{dx}=7(4 \frac{d}{dx} (log⁡x)+\frac{d}{dx} (log⁡5e^{x^3}))\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}} \frac{d}{dx} (5e^{x^3}))\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.\frac{d}{dx} {x^3})\)
\(\frac{dy}{dx}=7(\frac{4}{x}+\frac{1}{5e^{x^3}}.5e^{x^3}.3x^2)\)
\(\frac{dy}{dx}=7(\frac{4}{x}+3x^2)\)
∴\(\frac{dy}{dx}=\frac{7(4+3x^3)}{x}\)

9. Differentiate 2ex4 log⁡x w.r.t x.
a) \(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x^2}\)
b) \(\frac{e^{x^4} (4x^4 log⁡x+1)}{x}\)
c) \(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x}\)
d) –\(\frac{2e^{x^4} (4x^4 log⁡x+1)}{x}\)
View Answer

Answer: c
Explanation: Consider y=2ex4 log⁡x
\(\frac{dy}{dx}\)=\(\frac{d}{dx}\) (2ex4 log⁡x)
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=2(log⁡x \(\frac{d}{dx} (e^{x^4})+e^{x^4}\frac{d}{dx}\) (log⁡x))
\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}\frac{d}{dx} {x^4}+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2(log⁡x.e^{x^4}.4x^3+e^{x^4}.\frac{1}{x})\)
\(\frac{dy}{dx}\)=\(2e^{x^4} (4x^3 log⁡x+\frac{1}{x})\)
∴\(\frac{dy}{dx}=\frac{(2e^{x^4} (4x^4 log⁡x+1))}{x}\)
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10. Differentiate \(log⁡(cos⁡(sin⁡(e^{x^3})))\) w.r.t x.
a) –\(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
b) \(3x^2 \,e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
c) –\(3e^{x^3} \,cos⁡e^{x^3} \,cos⁡(sin⁡e^{x^3})\)
d) –\(x^2 e^{x^3} \,cos⁡e^{x^3} \,tan⁡(sin⁡e^{x^3})\)
View Answer

Answer: a
Explanation: Consider y=\(log⁡(cos⁡(sin⁡(e^{x^3})))\)
Differentiating w.r.t x by using chain rule, we get
\(\frac{dy}{dx}\)=\(\frac{d}{dx} (log⁡(cos⁡(sin⁡(e^{x^3}))))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} \frac{d}{dx} (cos⁡(sin⁡e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}) \frac{d}{dx}(sin⁡e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3}) \frac{d}{dx} (e^{x^3})))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3})(cos⁡e^{x^3})(e^{x^3}) \frac{d}{dx} {x^3}))\)
\(\frac{dy}{dx}\)=\((\frac{1}{cos⁡(sin⁡e^{x^3})} (-sin⁡(sin⁡e^{x^3}).cos⁡e^{x^3} .e^{x^3}.3x^2)\)
\(\frac{dy}{dx}\)=-\((\frac{3x^2 e^{x^3} cos⁡e^{x^3} sin⁡(sin⁡e^{x^3})}{cos⁡(sin⁡e^{x^3})})\)
\(\frac{dy}{dx}\)=-\(3x^2 e^{x^3} cos⁡e^{x^3} tan⁡(sin⁡e^{x^3})\)

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

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter