This set of Engineering Mathematics Interview Questions and Answers for freshers 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

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Explanation:

Leibniz rule is

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

Explanation: Leibniz rule is

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 xy_{3} + x^{2}y_{2} + x^{3}y_{0} = 0 then order of its nth differential equation is,

a) n

b) n+1

c) n+2

d) n+3

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Explanation:

1. If we differentiae this equation n times then terms comes in nth order differential equation is yn+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}+ x

^{2}y

_{n+2}+ 2nxy

_{n+1}+ 2n(n-1)y

_{n}+ x

^{3}y

_{n}+ 3nx

^{2}y

_{n-1}+ 6n(n-1)xy

_{n-2}+ 6n(n-1)(n-2)y

_{n-3}= 0

Hence order of differential equlation becomes n+3.

4. Find nth derivative of x^{2}y_{2} + xy_{1} + y = 0

a) x^{2}y_{n+2} + (2n+1)xy_{n+1} + (n^{2}+1)y_{n} = 0

b) x^{2}y_{n+2} + nxy_{n+1} + (n^{2}+1)y_{n} = 0

c) x^{2}y_{n+2} + (2n+1)xy_{n+1} + n^{2}y_{n} = 0

d) x^{2}y_{n+2} + 2nxy_{n+1} + n^{2}y_{n} = 0

View Answer

Explanation: x

^{2}y

_{2}+ xy

_{1}+ y = 0

By Leibniz theorem

x^{2}y_{n+2} + n(2x)(y_{n+1}) + n(n-1)(2)(y_{n}) + xy_{n+1} + ny_{n} + y_{n}= 0

x^{2}y_{n+2} + xy_{n+1}(2n+1) + y_{n}(n^{2}+1) = 0.

5. The nth derivative of x^{2}y_{2} + (1-x^{2})y_{1} + xy = 0 is,

a) x^{2}y_{n+2} + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0

b) x^{2}y_{n+2} + y_{n+1}(2nx-x^{2}) + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0

c) x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + y_{n}(2n^{2}-2n-2nx+x)-y_{n – 1}(2n^{2}-3n)=0

d) x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + xy_{n}2n^{2}-y_{n – 1}(2n^{2}-3n)=0

View Answer

Explanation: x^{2}y_{2} + (1-x^{2})y_{1} + xy = 0

Differenetiating n times by Leibniz Rule

x^{2}y_{n+2} + 2nxy_{n+1} + 2n(n-1)y_{n} + (1-x^{2})y_{n+1} – 2nxy_{n} – 2n(n-1)y_{n-1 }+ xy_{n} + ny_{n-1} = 0

x^{2}y_{n+2} + y_{n+1}(2nx+1-x^{2}) + y_{n}(2n^{2}-2n-2nx+x) – y_{n-1}(2n^{2}-3n)=0.

6. Find nth derivative of x^{n}Sin(nx)

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7. If y(x) = tan^{-1}x 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

Explanation:

Y = tan

^{-1}x

Y

_{1}= 1/(1+x

^{2})

(1+x

^{2})y

_{1}= 1

By Leibniz Rule,

(1+x

^{2})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^{-1}x, then

a) (1-x^{2})y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1)

b) x^{2}y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1)

c) (1-x^{2})y_{n+2} – 2nxy_{n+1} = ny_{n}(2n-1)

d) (1-x^{2})y_{n+2} – xy_{n+1}(2n-1) +ny_{n}(2n-1)=0

View Answer

Explanation: Y = sin

^{-1}x

Differentiating it

Y1 = 1/√(1-x

^{2})

(1-x^{2})(y1)^{2}= 1

Again Differentiating we get

(1-x^{2})2y_{1}y_{2} – 2x(y1)^{2} = 0

(1-x^{2})y_{2} = xy_{1}

By Leibniz Rule, Diff it n times,

(1-x^{2})y_{n+2} – 2xny_{n+1} – 2n(n-1)y_{n} = xy_{n+1} + ny_{n}

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2(n-1)+1)

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2n-1).

9. If y = cos^{-1}x, 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

Explanation: y = cos

^{-1}x

Differentiating it

Y1 = 1/√(1-x^{2} )

(1-x^{2})(y1)^{2}= 1

Again Differentiating we get

(1-x^{2})2y_{1}y_{2} – 2x(y1)^{2} = 0

(1-x^{2})y_{2} = xy_{1}

By Leibniz Rule, Diff it n times,

(1-x^{2})y_{n+2} – 2xny_{n+1} – 2n(n-1)y_{n} = xy_{n+1} + ny_{n}

(1-x^{2}) y_{n+2} – xy_{n+1}(2n-1) = ny_{n}(2(n-1)+1)

(1-x^{2}) 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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