Engineering Mathematics MCQ Questions and Answers Part 20

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

Answer: a
Explanation:
\(\frac{d(uvw)}{dx}=u\frac{d(uvw)}{dx}+vw\frac{d(u)}{dx}=u[v\frac{d(w)}{dx}+w\frac{d(v)}{dx}]+vw\frac{d(u)}{dx}\)=u’vw+uv’w+uvw’

2. If \(y=\frac{sin(x)e^x}{cos^2(x)}\), find ^dy⁄_dx .
a) Sec² (x) e^x [1 + Tan(x)] + e^x Tan(x)Sec(x)
b) Sec² (x) e^x [Sec(x) + Tan(x)] + e^x Tan(x)Sec(x)
c) Sec² (x) e^2x [Sec(x) + Tan(x)] + e^x Tan(x)Sec(x)
d) Sec(x) e^x [Sec(x) + Tan(x)] + e^x Tan(x)Sec(x)

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

3. Value of ^d⁄_dx⁡ [(1 + xe^x}{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) \(\frac{(1-Cos(x))(1+x) e^x – Sin(x)(1+xe^x)}{[1-Cos (x)]^2}\)

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

4. Find the derivative of Sin(x)Tan(x) w.r.t e^x 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) \(\frac{Sin(x)(1+Sec^2 (x))}{e^x (2+Tan^2 (x)+Tan(x))}\)

Answer: c
Explanation:
\(\frac{dSin(x)Tan(x)}{de^x Tan(x)} = \frac{dSin(x)Tan(x))}{dx} \frac{dx}{de^x Tan(x)}\)
=\(\frac{dSin(x)Tan(x)}{dx} \frac{1}{\frac{de^x Tan(x)}{dx}}\)
=\(\frac{[Sin(x)Sec^2 (x)+Sin(x)]}{e^x (Sec^2 (x)+Tan(x))}\)
=\(\frac{Sin(x)(1+Sec^2 (x))}{e^x (1+Tan^2 (x)+Tan(x))}\)

5. 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) \(2nTan^{n-1} xSec^2 x+nx^{n-1} Sec^2 x^n+1/(1+x^2)+nSec^2 (nx)\)

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

6. Evaluate d/dx x^x ln(x)
a) x^(x-1) + x^2x ln⁡(x) + x^x [ln⁡(x)]²
b) x^(x-1) + x^x ln⁡(x) + x^x [ln⁡(x)]²
c) x^(x-1) + x^x ln⁡(x) + x^x ln⁡(x)
d) x^x + x^x ln⁡(x) + x^x [ln⁡(x)]²

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

7. 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

Answer: b
Explanation:
\(tan^{-1}\frac{cos(x)-sin(x)}{cos(x)+sin(x)}=tan^{-1}tan(x)=x\)
Hence
\(\frac{d}{dx}tan^{-1}\frac{cos(x)-sin(x)}{cos(x)+sin(x)}=1\)

8. If y = Tan(x)^Tan(x) then ^dy⁄_dx⁡ = ?
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)

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

9. Evaluate ^d⁄_dx 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)

Answer: b
Explanation: ^d⁄_dx Cot(x)Cosec(x) = -Cosec³ (x) – Cosec² (x)Cot(x).(By multiplication rule)

10. 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) \(\frac{[2xSin(x)+x^2 Cos(x)]}{+Cosec(x)+Sec^2 (x)Cosec(x)}\)

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

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Engineering Mathematics Questions and Answers Part-20

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