# Engineering Mathematics Questions and Answers – The nth Derivative of Some Elementary Functions – 2

This set of Engineering Mathematics Interview Questions and Answers focuses on “The nth Derivative of Some Elementary Functions – 2”.

1. nth derivative of Sinh(x) is
a) 0.5(ex – e-x)
b) 0.5(e-x – ex)
c) 0.5(ex – (-1)n e-x)
d) 0.5((-1)-n e-x -ex)
View Answer

Answer: c
Explanation: Y = Sinh(x)
Y = 0.5[ex – e-x].
y1 = 0.5 [ex – (-1)e-x].
y2 = 0.5 [ex – (-1)2 e-x].
Similarly,
yn = 0.5 [ex – (-1)n e-x].

2. If y=log⁡(x(x2 – 1)), then nth derivative of y is ?
a) (-1)(n-1) (n-1)!(x(-n) + (x-1)(-n) + (x+1)(-n))
b) (-1)n (n)! (x(-n-1) + (x-1)(-n-1) + (x+1)(-n-1))
c) (-1)(n+1) (n+1)!(x(-n) + (x-1)(-n) + (x+1)(-n))
d) (-1)n(n)! (x(-n-1) + (x-1)(-n+1) + (x+1)(-n+1))
View Answer

Answer: a
Explanation: Y=log(x) – log(x2 – 1)
y1 = x(-1)-2x/(x2-1)
y1 = x(-1)-(x-1)(-1) + (x+1)(-1)
yn = (-1)(n-1) (n-1)!(x(-n)-(x-1)(-n) + (x+1)(-n)).

3. If x = a(Cos(t) + t2) and y = a(Sin(t) + t2 + t3) then dy/dx equals to
a) (Cos(t) + 3t2 + 2t) / (-Sin(t) + 2t)
b) (Sin(t) + 3t2 + 2t) / (-Cos(t) + 2t)
c) (Sin(t) + 3t2 + 2t) / (Cos(t) + 2t)
d) (Cos(t) + 3t2 + 2t) / (Sin(t) + 2t)
View Answer

Answer: a
Explanation: dx/dt = a(-Sin(t) + 2t)
dy/dt = a(Cos(t) + 2t + 3t2)
Then,
dy/dx = (Cos(t) + 3t2+2t)/(-Sin(t) + 2t).
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4. If y=tan(-1)⁡(x) , then which one is correct ?
a) y3 + y12 + 4xy2 y1=0
b) y3 + y12 + xy2 y1=0
c) y3 + 2y12 + xy2 y1=0
d) y2 + 2y12 + 4xy2 y1=0
View Answer

Answer: d
Explanation: y=tan-1⁡(x)
$$y_1 = \frac{1}{1+x^2}$$
$$y_2 = \frac{-2x}{(1+x^2)^2}$$
$$y_3 = -2[\frac{(1+x^2)^2-4x^2 (1+x^2)}{(1+x^2)^4}]$$
$$y_3 = -2[\frac{1}{1+(x^2)^2}-\frac{(4x^2)}{(1+x^2)^3}]$$
$$y_3+2y_1^2 + 4xy_2 y_1$$=0

5. What is the value of $$\frac{d^n (x^m)}{dx^n}$$ for m<n, m=n, m>n?
a) 0, n!, mPn x(m-n)
b) mPn x(m-n), n!, 0
c) 0, n!, mCn x(m-n)
d) mCn x(m-n), n!, 0
View Answer

Answer: a
Explanation: For, m > n
$$\frac{d^n (x^m)}{dx^n} = m \frac{d^{n-1} (x^{m-1})}{dx^{n-1}}=m(m-1)\frac{d^{n-2} (x^{m-2})}{dx^{n-2}}$$=……..
Since m>n, m-n=0 hence this cycle will moves upto (m-n) times and at last
$$\frac{d^n (x^m)}{dx^n}=m(m-1)(m-2)….(m-(n-1)) x^{m-n}$$
Hence,
$$\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}$$ ………. (1)
For m=n, from equation 1,
$$\frac{d^n (x^n)}{dx^n}=n_{P_n} x^{(n-n)}=n!$$
From m<n, from equation 1,
$$\frac{d^n (x^m)}{dx^n}=m_{P_n} x^{(m-n)}=0$$ (Because, $$m_{P_n}=0$$ for m<n)
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6. Which of the following is true
a) Value of $$\frac{d^m (Sin⁡(nx))}{dx^m}$$ is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π2 and n<0
b) Value of $$\frac{d^m (Sin⁡(nx))}{dx^m}$$ is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π2 and n>0
c) Value of $$\frac{d^m (Sin⁡(nx))}{dx^m}$$ is always positive for m=0, 1, 4, 5, 8, 9… for 0 < nx < π2 and n>0
d) Value of $$\frac{d^m (Sin⁡(nx))}{dx^m}$$ is always positive for m=2, 3, 6, 7, 10, 11… for 0 < nx < π2 and n<0
View Answer

Answer: c
Explanation: Here,
$$\frac{d(Sin⁡(nx))}{dx} = n Cos(nx)$$ …………….(m=1)
$$\frac{d^2 (Sin⁡(nx))}{dx^2} = -n^2 Sin(nx)$$ …..(m=2)
$$\frac{d^3 (Sin⁡(nx))}{dx^3} = -n^3 Cos(nx)$$ …..(m=3)
$$\frac{d^4 (Sin⁡(nx))}{dx^4} = n^4 Sin(nx)$$ ……(m=4)
So the value of $$\frac{d^m (Sin⁡(nx))}{dx^m} = \begin{cases}n^m Cos(nx) \,\,\, m=1,5,9,….\\-n^m Sin(nx) \,\,\, m=2,6,10…\\-n^m Cos(nx)\,\,\,m=3,7,11…..\\n^m Sin(nx)\,\,\,m=4,8,12….\end{cases}$$
Hence, for n>0 and 0<nx<(π)/2 only $$\frac{d^m (Sin⁡(nx))}{dx^m}$$ is positive only when m = 1,4,5,8,9,….. otherwise negative.

7. If nth derivative of eax sin⁡(bx+c) cos⁡(bx+c) is, eax rn sin⁡(bx+c+2) cos⁡(bx+c+2) then,
a) r = $$\sqrt{a^2+b^2}, \alpha=tan^{-1}\frac{⁡b}{a}$$
b) r = $$\sqrt{a^2+4b^2}, \alpha=tan^{-1}⁡\frac{2b}{a}$$
c) r = $$\sqrt{a^2+8b^2}, \alpha=tan^{-1}⁡\frac{4b}{a}$$
d) r = $$\sqrt{a^2+16b^2}, \alpha=tan^{-1}\frac{⁡4b}{a}$$
View Answer

Answer: b
Explanation: y = eax sin⁡(bx+c) cos⁡(bx+c)
y = eax sin⁡2(bx+c)/2
yn = eax rn sin⁡(2(bx+c+nα/2))/2
yn = eax rn sin⁡(bx+c+nα/2) cos⁡(bx+c+nα/2)
where
r = $$\sqrt{a^2+4b^2}$$, α = tan-1⁡2b/a.
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8. If y=x4x2-1, then?
a) 0.5*(-1)n (n-1)! [(x-1)-n-1 + (x+1)-n-1]
b) 0.5*(-1)n (n-1)! [x– n-1 + (x-1)-n-1 + (x+1)-n-1]
c) 0.5*(-1)n (n-1)! [(x-1)-n + (x+1)-n)]
d) 0.5*(-1)n (n-1)! [x-n + (x-1)-n + (x+1)-n]
View Answer

Answer: a
Explanation: $$y = x^2+\frac{x^2}{x^2-1}$$
$$y = x^2+\frac{1}{x^2-1}$$+1
$$y = x^2+1+\frac{1}{(x-1)(x+1)}$$
$$y = x^2+1+0.5[\frac{1}{(x-1)}-\frac{1}{(x+1)}]$$
$$y_n=0.5*(-1)^n (n-1)![(x-1)^{-n-1}+(x+1)^{-n-1}]$$

9. If y=sin(-1)⁡(x) then select the true statement.
a) y2 = xy13
b) y3 = xy23
c) y2 = xy12
d) y3 = xy12
View Answer

Answer: a
Explanation: y=sin-1⁡(x)
$$y_1=\frac{1}{\sqrt{1-x^2}}$$
$$y_2=\frac{x}{(1-x^2)^{3/2}}$$
$$y_2=\frac{x}{(1-x^2)^{3/2}} = xy_1^3$$
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10. nth derivative of y = sin2x cos3x is
a) 18 cos⁡(x + 2) –116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2)
b) 18 sin⁡(x+2) –116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2)
c) 18 cos⁡(x+2) –116 5n sin⁡(x + 2) – 116 3n sin⁡(3x + 2)
d) 18 sin⁡(x + 2) –116 5n sin⁡(x + 2) – 116 3n sin⁡(3x + 2)
View Answer

Answer: a
Explanation: y = sin2x cos2x cos(x)
y = 14 sin22x cos2x cos(x)
y = 18 (2sin22x) cos(x)
y = 18 (1 – cos4x) cos(x)
y = 18 (1 – cos4x) cos(x)
y = 18 cos(x) – 18 cos4x cos(x)
y = 18 cos(x) – 116 (cos5x + cos(3x))
Now, nth derivative is
yn = 18 cos⁡(x + 2) – 116 5n cos⁡(x + 2) – 116 3n cos⁡(3x + 2).

11. If In=enxTan(x), and $$\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}$$ = c (1+x2)$$\frac{d^2}{dx^2} (\frac{1}{1+x^2})$$, Then value of ‘c’ equals to
a) 1
b) 2
c) 3
d) 4
View Answer

Answer: a
Explanation: In=enxTan(x)
⇨In+1=nexnTan(x) + exn/(1+x2)
⇨In+1=nIn + exn/(1+x2)
⇨In+2=nIn+1 + $$\frac{ne^nx}{1+x^2} – \frac{2xe^nx}{(1+x^2)^2}$$
⇨In+2=nIn+1 + n[In+1-nIn] – $$\frac{2xe^nx}{(1+x^2)^2}$$
⇨$$\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}=(1+x^2)\frac{d^2}{dx^2} (\frac{1}{1+x^2})$$, Hence c=1.

12. Nth derivative of $$\frac{1}{(1+x^2)}$$?
a) (-1)nn! r-n-1 Sin(n+1)ϑ
b) (-1)n(n)! r-n-1 Sin(-n-1)ϑ
c) (-1)n+1(n+1)! r-n-1 Sin(n+1)ϑ
d) (-1)n+1(n+1)! r-n-1 Sin(-n-1)ϑ
View Answer

Answer: a
Explanation: Here,
Y = $$\frac{1}{(x+i)(x-i)}$$
Y = $$\frac{1}{2i(x-i)} – \frac{1}{2i(x+i)}$$
$$y_n = \frac{-1^n n!}{2i}[(x-i)^{-n-1}-(x+i)^{-n-1}]$$
put x=rcos ϑ and y=rsinϑ, we get
$$y_n=\frac{(-1)^n n!}{2i}[(re^{-iθ})^{-n-1}-(re^{iθ})^{-n-1}]$$ (By Eulers Identity)
$$y_n=\frac{(-1)^n n!r^{-n-1}}{2i}[e^{iθ(n+1)}-e^{-iθ(n+1)}]$$
$$y_n=(-1)^n n!r^{-n-1} Sin((n+1)ϑ)$$

13. Find nth derivative of y = Sin(x) Cos3(x)
a) (1/4) 2nSin(2x+nπ/2) + (1/8) 4nSin(4x+nπ/2)
b) (1/4) 2nCos(2x+nπ/2) + (1/8) 4nSin(4x+nπ/2)
c) (1/4) 2nSin(2x+nπ/2) + (1/8) 4nCos(4x+nπ/2)
d) (1/4) 2nCos(2x+nπ/2) + (1/8) 4nCos(4x+nπ/2)
View Answer

Answer: a
Explanation: y = Sin(x) Cos3(x)
y = (1/2)[2Sin(x) Cos(x)] Cos2(x)
y = (1/4)Sin(2x)(Cos(2x)+1)
y = (1/4)[Sin(2x)Cos(2x)+Sin(2x)] y = (1/8)Sin(4x) + (1/4)Sin(2x)
Hence nth derivative of y is
Yn = (1/4) 2nSin(2x+nπ/2) + (1/8) 4nSin(4x+nπ/2)

14. If nth derivative of y = $$\frac{x}{(x+1)(x+2)}$$ is yn = a(-1)n+! n! (x+1)-n-1 + b(-1)nn!(x+2)-n-1 then find the value of a and b.
a) -1, -2
b) 2, 1
c) 1, 2
d) -2, -1
View Answer

Answer: c
Explanation: y = $$\frac{x}{(x+1)(x+2)}$$
y = $$\frac{(-1)}{(x+1)} + \frac{2}{(x+2)}$$
y = (-1)(-1)nn!(x+1)-n-1 + 2(-1)nn!(x+2)-n-1
y = (-1)n+1n!(x+1)-n-1 + 2(-1)nn!(x+2)-n-1
Hece, a=1 and b=2.

Sanfoundry Global Education & Learning Series – Engineering Mathematics.

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