Engineering Mathematics Questions and Answers - Leibniz Rule - 3

This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Leibniz Rule – 3”.

1. Leibniz rule gives the
a) Nth derivative of addition of two function
b) Nth derivative of division of two functions
c) Nth derivative of multiplication of two functions
d) Nth derivative of subtraction of two function
View Answer

Answer: c
Explanation: Leibniz rule is
\(\frac{d^n}{dx^n}(uv)\)
\(= n_{C_0} u_n v+ n_{C_1} u_{(n-1)} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n\)
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v.
.

2. Leibniz theorem is applicable if n is a
a) Rational Number
b) Negative Integer
c) Positive Integer
d) Decimal Number
View Answer

Answer: c
Explanation: Leibniz rule is
\(\frac{d^n}{dx^n}(uv)\)
\(= n_{C_0} u_n v+ n_{C_1} u_{n-1} v_1+ n_{C_2} u_{n-2} v_2+…+ n_{C_r} u_{n-r} v_r+…+ n_{C_n} u_0 v_n\)
For all n > 0, i.e n should be positive
Hence Leibniz theorem gives nth derivative of multiplication of two functions u and v if n is a positive integer.

3. If nth derivative of xy₃ + x²y₂ + x³y₀ = 0 then order of its nth differential equation is,
a) n
b) n+1
c) n+2
d) n+3
View Answer

Answer: d
Explanation:
1. If we differentiate this equation n times then terms comes in nth order differential equation is y_n+3, y_n+2, y_n+1, y_n, y_n-1, y_n-2, y_n-3. Hence order of differential equation becomes n+3.
2. By Leibniz rule differentiating it n times, we get
Xy_n+3 + ny_n+2 + x²y_n+2 + 2nxy_n+1 + 2n(n-1)y_n + x³y_n + 3nx²y_n-1 + 6n(n-1)xy_n-2 + 6n(n-1)(n-2)y_n-3 = 0
Hence order of differential equation becomes n+3.

4. Find nth derivative of x²y₂ + xy₁ + y = 0
a) x²y_n+2 + (2n+1)xy_n+1 + (n²+1)y_n = 0
b) x²y_n+2 + nxy_n+1 + (n²+1)y_n = 0
c) x²y_n+2 + (2n+1)xy_n+1 + n²y_n = 0
d) x²y_n+2 + 2nxy_n+1 + n²y_n = 0
View Answer

Answer: a
Explanation: x²y₂ + xy₁ + y = 0
By Leibniz theorem
x²y_n+2 + n(2x)(y_n+1) + n(n-1)(2)(y_n) + xy_n+1 + ny_n + y_n= 0

x²y_n+2 + xy_n+1(2n+1) + y_n(n²+1) = 0.

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5. The nth derivative of x²y₂ + (1-x²)y₁ + xy = 0 is,
a) x²y_n+2 + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0
b) x²y_n+2 + y_n+1(2nx-x²) + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0
c) x²y_n+2 + y_n+1(2nx+1-x²) + y_n(2n²-2n-2nx+x)-y_{n – 1}(2n²-3n)=0
d) x²y_n+2 + y_n+1(2nx+1-x²) + xy_n2n²-y_{n – 1}(2n²-3n)=0
View Answer

Answer: c
Explanation: x²y₂ + (1-x²)y₁ + xy = 0
Differentiating n times by Leibniz Rule
x²y_n+2 + 2nxy_n+1 + 2n(n-1)y_n + (1-x²)y_n+1 – 2nxy_n – 2n(n-1)y_n-1+ xy_n + ny_n-1 = 0
x²y_n+2 + y_n+1(2nx+1-x²) + y_n(2n²-2n-2nx+x) – y_n-1(2n²-3n)=0.

6. Find nth derivative of xⁿSin(nx)
a) \(n!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx)\)
\(-n_{C_3}n_{P_{n-3} }x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2)\)
b) \(Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx)\)
\(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+n!x^nn^nSin(nx+nπ/2)\)
c) \(nSin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx)\)
\(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+nx^nn^nSin(nx+nπ/2)\)
d) \((n-1)!Sin(nx) + n_{C_1} n_{P_{n-1}}x^{n-1}nCos(nx) – n_{C_2}n_{P_{n-2}}x^{n-2}n^2Sin(nx)\)
\(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^{n-1}Sin(nx+nπ/2)\)
View Answer

Answer: a
Explanation: Y = xⁿSin(nx)
By Leibniz Rule , put u = xⁿ and v = Sin(nx), we get
\(n!Sin(nx) + n_{C_1} n_(P_{n-1})x^{n-1}nCos(nx) – n_{C_2}n_(P_{n-2})x^{n-2}n^2Sin(nx)\)
\(-n_{C_3}n_{P_{n-3}}x^{n-3}n^3Sin(nx)+n_{C_3}n_{P_{n-4}}x^{n-4}n^4Cos(nx)+..+x^nn^nSin(nx+nπ/2)\)

7. If y(x) = tan^-1x then
a) (y_n+1)(0) = (n-1)(y_n-1)0
b) (y_n+1)(0) = n(n-1)(y_n-1)0
c) (y_n+1)(0) = -(n-1)(y_n-1)0
d) (y_n+1)(0) = -n(n-1)(y_n-1)0
View Answer

Answer: d
Explanation: Y = tan^-1x
Y₁ = 1/(1+x²)
(1+x²)y₁ = 1
By Leibniz Rule,
(1+x²)y_n+1 + 2nxy_n + n(n-1)y_n-1 = 0
Put x=0, gives
→ (y_n+1)(0) = -n(n-1)(y_n-1)(0).

8. If y = sin^-1x, then
a) (1-x²)y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
b) x²y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
c) (1-x²)y_n+2 – 2nxy_n+1 = ny_n(2n-1)
d) (1-x²)y_n+2 – xy_n+1(2n-1) +ny_n(2n-1)=0
View Answer

Answer: a
Explanation: Y = sin^-1x
Differentiating it
Y₁ = \(\frac{1}{\sqrt{1-x^2}}\)
(1-x²)(y₁)²= 1
Again Differentiating we get
(1-x²)2y₁y₂ – 2x(y₁)² = 0
(1-x²)y₂ = xy₁
By Leibniz Rule, Diff it n times,
(1-x²)y_n+2 – 2xny_n+1 – 2n(n-1)y_n = xy_n+1 + ny_n
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2(n-1)+1)
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2n-1).

9. If y = cos^-1x, then
a) (y_n+2)(0) = -n(2n-1) y_n(0)
b) (y_n+2)(0) = n(2n-1) y_n(0)
c) (y_n+2)(0) = n(n-2) y_n(0)
d) (y_n+2)(0) = n(n-3) y_n(0)
View Answer

Answer: b
Explanation: y = cos^-1x
Differentiating it
Y₁ = \(\frac{1}{\sqrt{1-x^2}}\)
(1-x²)(y₁)²= 1
Again Differentiating we get
(1-x²)2y₁y₂ – 2x(y₁)² = 0
(1-x²)y₂ = xy₁
By Leibniz Rule, Diff it n times,
(1-x²)y_n+2 – 2xny_n+1 – 2n(n-1)y_n = xy_n+1 + ny_n
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2(n-1)+1)
(1-x²) y_n+2 – xy_n+1(2n-1) = ny_n(2n-1)
At x=0, we get
(y_n+2)(0) = n(2n-1) y_n(0).

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