Engineering Mathematics Questions and Answers – Limits and Derivatives of Several Variables – 4

This set of Engineering Mathematics online quiz focuses on “Limits and Derivatives of Several Variables – 4”.

1. \(\frac{d(uvw)}{dx}\) is where u ,v, w are the functions of x
a) u’vw + uv’w + uvw’
b) uvw + uv’w’ + u’v’w’
c) u’v’w + uv’w’ + u’vw’
d) uv’w’ + u’v’w’ + uvw
View Answer

Answer: a

2. \(\frac{d(\frac{u}{v})}{dx}\) is where u, v are the functions of x
a) v’u’ – uvv2
b) vu’ – uv’v2
c) vu – u’v’v2
d) 0
View Answer

Answer: b

3. If \(y=\frac{sin(x)e^x}{cos^2(x)}\), find dydx .
a) Sec2 (x) ex [1 + Tan(x)] + ex Tan(x)Sec(x)
b) Sec2 (x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
c) Sec2 (x) e2x [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
d) Sec(x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x)
View Answer

Answer: c
Explanation: \(y=\frac{sin(x)e^x}{cos^2(x)}\) = Tan(x)Sec(x) ex
dydx = Sec2 (x)Sec(x) ex + Sec2 (x)Tan(x) ex + ex Tan(x)Sec(x)
dydx = Sec2 (x) ex [Sec(x) + Tan(x)] + ex Tan(x)Sec(x).

4. Value of ddx⁡ [(1 + xex}{1-Cos(x))].
a) \(\frac{(1-Sin(x))(1+x) e^x + Cos(x)(1+xe^x)}{[1-Cos (x)]^2}\)
b) \(\frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^4}\)
c) \(\frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^2}\)
d) \(\frac{(1-Cos(x))(1+x) e^x – Sin(x)(1+xe^x)}{[1-Cos (x)]^2}\)
View Answer

Answer: c
\(\frac{d}{dx⁡} (1+xe^x)/(1-Cos(x))\)
\(\frac{d}{dx⁡} (1+xe^x)/(1-Cos(x)) = \frac{(1-Cos(x))(1+x) e^x + Sin(x)(1+xe^x)}{[1-Cos (x)]^2}\)

5. Find the derivative of Sin(x)Tan(x) w.r.t ex Tan(x)
a) \(\frac{Sin(x)(1+Sec^4 (x))}{e^x (1+Tan^2 (x)+Tan(x))}\)
b) \(\frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^4 (x)+Tan(x))}\)
c) \(\frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^2 (x)+Tan(x))}\)
d) \(\frac{Sin(x)(1+Sec^2 (x))}{e^x (2+Tan^2 (x)+Tan(x))}\)
View Answer

Answer: c
\(\frac{dSin(x)Tan(x)}{de^x Tan(x)} = \frac{dSin(x)Tan(x))}{dx} \frac{dx}{de^x Tan(x)}\)
=\(\frac{dSin(x)Tan(x)}{dx} \frac{1}{\frac{de^x Tan(x)}{dx}}\)
=\(\frac{[Sin(x)Sec^2 (x)+Sin(x)]}{e^x (Sec^2 (x)+Tan(x))}\)
=\(\frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^2 (x)+Tan(x))}\)
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6. Evaluate \(\frac{d[Tan^n (x)+Tanx^n+Tan^{-1} x+Tan(nx)}{dx}]\) is
a) \(nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nTan(nx)Sec^2 (nx)\)
b) \(nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nSec^2 (nx)\)
c) \(nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1-x^2)+nSec^2 (nx)\)
d) \(2nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nSec^2 (nx)\)
View Answer

Answer: b
Explanation: \(\frac{d[Tan^n (x)+Tanx^n+Tan^{-1} x+Tan(nx)}{dx}]\)
\(=nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+\frac{1}{1+x^2}+nSec^2 (nx)\).

7. Evaluate d/dx xx ln(x)
a) x(x-1) + x2x ln⁡(x) + xx [ln⁡(x)]2
b) x(x-1) + xx ln⁡(x) + xx [ln⁡(x)]2
c) x(x-1) + xx ln⁡(x) + xx ln⁡(x)
d) xx + xx ln⁡(x) + xx [ln⁡(x)]2
View Answer

Answer: b
\(\frac{dx^x ln⁡(x)}{dx}=\frac{x^x dln(x)}{dx}+\frac{ln⁡(x)dx^x}{dx}\)
Y = xx
Taking log on both side.
ln(y) = xln(x)
Differntiating both sides
\(\frac{1}{y} \frac{dy}{dx}=1+ln⁡(x)\)
\(\frac{dy}{dx}\)=xx (1+ln⁡(x))
\(\frac{dx^x ln⁡(x)}{dx}=x^{x-1}+x^x ln⁡(x) (1+ln⁡(x) )=x^{x-1}+x^x ln⁡(x)+x^x [ln⁡(x)]^2\)

8. Evaluate the differentiation of \(tan^{-1}\frac{cos(x)-sin(x)}{cos(x)+sin(x)}\)
a) tan-1⁡x
b) 1
c) 0
d) -1
View Answer

Answer: b

9. If y = Tan(x)Tan(x) then dydx⁡ = ?
a) Tan(x) [1 + lnTan(x)] Tan(x)Tan(x)
b) Tan2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
c) Sec2 (x) [1 + lnTan(x)] Tan(x)Tan(x)
d) Sec(x) [1 + lnTan(x)] Tan(x)Tan(x)
View Answer

Answer: c
Explanation: y = Tan(x)Tan(x)
Taking ln on both side
ln y = Tan(x)lnTan(x)
Differentiating w.r.t x
\(\frac{1}{y} \frac{dy}{dx} = \frac{Tan(x)Sec^2 (x)}{Tan(x)}+Sec^2 (x)lnTan(x)\)
\(\frac{1}{y} \frac{dy}{dx} = Sec^2 (x)[1+lnTan(x)]\)
\(\frac{dy}{dx}=Sec^2 (x)[1+lnTan(x)]Tan(x)^{Tan(x)}\)

10. Evaluate ddx Cot(x)Cosec(x)
a) -Cosec2 (x) – Cosec2 (x)Cot(x)
b) -Cosec3 (x) – Cosec2 (x)Cot(x)
c) -Cosec(x) – Cosec2 (x)Cot(x)
d) -Cosec3 (x) – Cosec(x)Cot2 (x)
View Answer

Answer: b
Explanation: ddx Cot(x)Cosec(x) = -Cosec3 (x) – Cosec2 (x)Cot(x).(By multiplication rule)

11. Evaluate differentiation of x2 Sin(x) w.r.t Tan(x)Cosec(x)
a) \(\frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)-Sec^2 (x)Cosec(x)}\)
b) \(\frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Cos(x)Sin(x)}\)
c) \(\frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Sec^2 (x)Cosec(x)}\)
d) \(\frac{[2xSin(x)+x^2 Cos(x)]}{+Cosec(x)+Sec^2 (x)Cosec(x)}\)
View Answer

Answer: c
\(\frac{dx^2 Sin(x)}{dTan(x)Csoec(x)}=\frac{dx^2 Sin(x)}{dx} \frac{dx}{dTan(x)Cosec(x)}\)
\(=\frac{[2xSin(x)+x^2 Cos(x)]}{-Cosec(x)+Sec^2 (x)Cosec(x)}\)

12. If z = ex Sin(Cos(x))Cos(Sin(x)) Then find dzdx
a) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))]
b) [exSin(Cos(x))Cos(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))]
c) [exCos(Cos(x))Sin(Sin(x))-exSin(x)Cos(Cos(x))Cos(Sin(x))-exCos(x)Sin(Cos(x))Sin(Sin(x))]
d) [exSin(Cos(x))Cos(Sin(x))-exCos(x)Cos(Cos(x))Cos(Sin(x))-exSin(x)Sin(Cos(x))Sin(Sin(x))]
View Answer

Answer: b
Explanation: dzdx = ddx ex Sin(Cos(x))Cos(Sin(x)) = [(ex Sin(Cos(x))Cos(Sin(x)) – ex Sin(x)Cos(Cos(x))Cos(Sin(x)) – ex Cos(x)Sin(Cos(x))Sin(Sin(x)))].

13. If F(x) = f(x)g(x)h(x) and F’(x) = 10F(x) and f’(x) = 10f(x) , g’(x) = 10g(x) and h’(x) = 10kh(x), then find value of k.
a) 0
b) 1
c) -1
d) 2
View Answer

Answer: c
Explanation: Given F(x) = f(x)g(x)h(x)
F’(x) = f’(x)g(x)h(x) + f(x)g’(x)h(x) + f(x)g(x)h’(x)
Putting value of F’(x), f’(x), g’(x), h’(x)
We get
10 = 10 + 10 + 10k
K = -1.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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