This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Leibniz Rule – 1”.

1. Let f(x) = sin(x) / 1+x^{2} . Let y^{(n)} denote the n^{th} derivative of f(x) at x = 0 then the value of y^{(100)} + 9900y^{(98)} is

a) 0

b) -1

c) 100

d) 1729

View Answer

Explanation:The key here is a simple manipulation and application of the Leibniz rule.

Rewriting the given function as

y(1 + x

^{2}) = sin(x)…….(1)

The Leibniz rule for two functions is given by

2. Let f(x) = ln(x)/x+1 and let y^{(n)} denote the n^{th} derivative of f(x) at x = 1 then the value of 2y^{(100)} + 100y^{(99)}

a) (99)!

b) (-99)!

c) (100)!

d) (-98)!

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Explanation:Assume f(x) = y

Rewrite the function as

y(x + 1 ) = ln(x)

Now differentiate both sides up to hundredth derivative in accordance to the Leibniz rule we have

3. Let f(x) = √1-x^{2} and let y^{(n)} denote the n^{th} derivative of f(x) at x = 0 then the value of 6y ^{(1)} y^{(2)} + 2y^{(3)} is

a) -998

b) 0

c) 998

d) -1

View Answer

Explanation:Assume f(x) = y

Rewriting the function as

y

^{2}= 1 – x

^{2}

Differentiating both sides of the equation up to the third derivative using leibniz rule we have

Now substituting x = 0 in both the equations and equating them yields

2y

^{(3)}y + 6y

^{(2)}y

^{(1)}= 0.

4. Let f(x) = tan(x) and let y^{(n)} denote the n^{th} derivative of f(x) then the value of y^{(9998879879789776)} is

a) 908090988

b) 0

c) 989

d) 1729

View Answer

Explanation:Assume y = f(x) and we also know that

Using (1) and the above equation one can conclude that

y

^{(2)}= 0

This gives the value of second derivative to be zero

Similarly for any even value of n all the odd derivatives of y in

the expression would have sin(x) as their coefficients and as the values of y

^{(0)}and y

^{(2)}are zero. Every even derivative of the tan(x) function has to be zero.

Thus, we have

y^{(9998879879789776)} = 0.

5. If the first and second derivatives at x = 0 of the function

were 2 and 3 then the value of the third derivative is

a) -3

b) 3

c) 2

d) 1

View Answer

Explanation: Assume

f(x) = y

Write the given function as

y(x

^{2}– x + 1) = cos(x)

Now applying Leibniz rule up to the third derivative we get

Substituting x = 0 gives

sin(0) = y^{(3)} + 9 -12

y^{(3)} = 12 – 9 = 3.

6. For the given function the values of first and second derivative at x = 0 are assumed as 0 and 1 respectively. Then the value of the third derivative could be

a) 54√2

b) 2√2

c) √2

d) Indeterminate

View Answer

Explanation:Rewriting the given function as

y

^{2}= x

^{3}+ x

^{7}

Now applying the Leibniz rule up to the third derivative we have

Equating both sides and substituting x = 1 we get

y

^{(1)}= 0

Now assumed in the question are the values y

^{(2)}= 1 and y(1) = √2

We also know

2√2 y^{(3)} = 3! + 210 = 216

Putting them in equation (1) we get

y^{(3)} = 54√2.

7. Let and let the n^{th} derivative at x = 0 be given by y^{(n)} Then the value of the expression for y^{(n)} is given by

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Explanation:

8. Let f(x) = e^{x} sinh(x) / x , let y^{(n)} denote the n^{th} derivative of f(x) at x = 0 then the expression for y^{(n)} is given by

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**Sanfoundry Global Education & Learning Series – Engineering Mathematics.**

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