This set of Engineering Mathematics Questions and Answers for Freshers focuses on “Leibniz Rule – 2”.

1. Let f(x) = e^{x} sin(x^{2}) ⁄ x Then the value of the fifth derivative at x = 0 is given by

a) 25

b) 21

c) 0

d) 5

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2. Let f(x) = e^{ex} assuming all the n^{th} derivatives at x =0 to be 1 the value of the (n + 1)^{th} derivative can be written as

a) e – 1 + 2^{n}

b) 0

c) 1

d) None

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Explanation: Assume y = f(x)

Taking ln(x) on both sides The function has to be written in the form ln(y) = e

^{y}

Now computing the first derivative yields

y

^{(1)}= y * e

^{x}

Now applying the Leibniz rule up to n

^{th}derivative we have

3. Let f(x) = √sin(x) and let y^{n} denote the n^{th} derivative of f(x) at x = 0 then the value of the expression 12y^{(5)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2} is given by

a) 0

b) 655

c) 999

d) 1729

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Explanation: Assume y = f(x)

Rewriting the function as y

^{2}= sin(x)

Now applying Leibniz rule up to the sixth derivative we have

(y

^{2})

^{(6)}= c

_{0}

^{6}y

^{(6)}y + c

_{1}

^{6}y

^{(5)}y(1) + ………+ c

_{6}

^{6}y

^{(6)}y

(y^{2})^{(6)} = 2 y^{(6)} y + 12 y^{(6)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2}

(sin(x))^{(6)} = -sin(x)

Now substituting x = 0 and observing that y(0)= 0 we have

sin(0) = 0 = 12 y^{(6)} y^{(1)} + 30 y^{(4)} y^{(2)} + 20 (y^{(3)})^{2}.

4. The fourth derivative of f(x) = sin(x)sinh(x) ⁄ x at x = 0 is given by

a) 0

b) ^{π}⁄_{2}

c) 45

d) 4

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5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is

a) 0

b) ^{π}⁄_{32}

c) (π)^{2}

d) cos(1)sinh(1)

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Explanation: Assume y = f(x)

Rewriting the part sinh(x)⁄x as infinite series we have

Now substituting x = 0 we have

y^{(3)} = 0.

6. Let f(x) = (x^{2} + x + 1)sinh(x) the (1097)^{th} derivative at x = 0 is

a) 1097

b) 1096

c) 0

d) 1202313

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7. The 7^{th} derivative of f(x) = (x^{3} + x^{2} + x + 1) sinh(x) at x = 0 is given by

a) 43

b) 7

c) 0

d) 34

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8. The (1071729)^{th} derivative of f(x) = (x^{6} + x^{4} + x^{2}) cosh(x) at x = 0 is given by

a) 0

b) 1071

c) 1729

d) ∞

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Explanation: Expanding cosh(x) into a Taylor series we have

Observe again, that the derivative in question is odd, and hence, only the odd powered terms contribute to the derivative at x = 0

Also note that, there are no odd powered terms and hence we can conclude that

The (1071729)^{th} derivative must be 0.

9. The (17291728)^{th} derivative of f(x) = (x^{2} + 1)tan^{-1} (x) at x = 0 is

a) 0

b) 1729

c) 1728

d) ∞

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Explanation: Expanding the tan

^{-1}(x) function into Taylor series we have

Now observe that the derivative in question is even. Hence, even terms are the only ones to contribute to the derivative at x = 0

Also note that there are no even powered terms in the function. One can conclude that the (17291728)^{th} derivative at x = 0 is 0.

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

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