Leibniz Rule - Engineering Mathematics Questions and Answers

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

1. Let f(x) = sin(x)/1+x². 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

Answer: a
Explanation:The key here is a simple manipulation and application of the Leibniz rule.
Rewriting the given function as
y(1 + x²) = sin(x)…….(1)
The Leibniz rule for two functions is given by
(uv)⁽ⁿ⁾=\(c_{0}^{n}u(v)^{(n)}+c_{1}^{n}u^{(1)}(v)^{(n-1)}+….+c_{n}^{n}u^{(n)}v\)
Differentiating expression (1) in accordance to Leibniz rule (upto the hundredth derivative) we have
(y(1+x²))⁽¹⁰⁰⁾ = \(c_{0}^{100}y^{(100)}(1+x^2)+c_{1}^{100}y^(99)(2x)+c_{2}^{100}y^(98)(2)+0….+0\)
(y(1+x²))=(sin(x))⁽¹⁰⁰⁾=sin(x)
Now substituting gives us
y⁽¹⁰⁰⁾+9900y⁽⁹⁸⁾=sin(0)=0
Hence, Option 0 is the required answer.

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)!
View Answer

Answer: b
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
(y(1+x))⁽¹⁰⁰⁾=\(c_0^{100}y^{(100)}(x+1)+c_1^{100}y^{(99)}+0+…..+0=(ln(x))^{(100)}\)
Using the n^th derivative for ln(x+a) as \(\frac{d^n(ln(x+a))}{dx^n}=\frac{(-1)^{n+1}\times(n-1)!}{(x+a)^n}\)
we have the right hand side as
(y(1+x))⁽¹⁰⁰⁾ = \(\frac{(-99)!}{x^{100}}\)
Now substituting x = 1 yields
2y⁽¹⁰⁰⁾ + y⁽⁹⁹⁾ = \(\frac{(-99)!}{1}\)
= -(99)!

3. Let f(x) = \(\sqrt{1-x^2}\) 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

Answer: b
Explanation:Assume f(x) = y
Rewriting the function as
y² = 1 – x²
Differentiating both sides of the equation up to the third derivative using leibniz rule we have
(y²)⁽³⁾=\(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3y^{(3)}y\)
(y²)⁽³⁾=\(2y^{(3)}y+6y^{(2)}y^{(1)}\)
(1-x²)⁽³⁾=0
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

Answer: b
Explanation:Assume y = f(x) and we also know that tan(x)=\(\frac{sin(x)}{cos(x)}\)
Rewriting the function as y(cos(x))=sin(x)
Now differentiating on both sides upto n^nt derivative we have
(y(cos(x))⁽ⁿ⁾=\(c_0^ny^{(n)}cos(x)-c_1^ny^{(n-1)}sin(x)+….+(cos(x))^{(n)}y\)
Now observe that y(0)=tan(0)=0….(1)
Now consider the second derivative at x=0 on both sides
(y(cos(x))⁽²⁾=\(c_0^2y^{(2)}cos(0)-c_1^2y^{(1)}sin(0)-c_2^2ycos(0)=(sin(x))^{(2)}=0\)
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 f(x)=\(\frac{cos(x)}{x^2-x+1}\) were 2 and 3 then the value of the third derivative is
a) -3
b) 3
c) 2
d) 1
View Answer

Answer: b
Explanation: Assume
f(x) = y
Write the given function as
y(x² – x + 1) = cos(x)
Now applying Leibniz rule up to the third derivative we get
\((y(x^2-x+1))^{(3)}=c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2)(cos(x))^{(3)}\)=sin(x)
Equating both sides we have
\(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3y^{(2)}(2x-1)-c_2^3y^{(1)}(2)\)
Now in the question it is assumed that the y⁽¹⁾=2 and y⁽²⁾=3
Substituting these values in (1) we have
\(sin(x) = c_0^3y^{(3)}(x^2-x+1)-c_1^3(3)(2x-1)-c_2^3(2)(2)\)
Substituting x = 0 gives
sin(0) = y⁽³⁾ + 9 -12
y⁽³⁾ = 12 – 9 = 3.

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6. For the given function f(x)=\(\sqrt{x^3+x^7}\) the values of first and second derivative at x = 1 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

Answer: a
Explanation:Rewriting the given function as
y² = x³ + x⁷
Now applying the Leibniz rule up to the third derivative we have
(y²)⁽³⁾=(x³+x⁷)⁽³⁾
\(c_0^3y^{(3)}y+c_1^3y^{(2)}y^{(1)}+c_2^3y^{(1)}y^{(2)}+c_3^3yy^{(3)}=3!+(7.6.5)x^4\)….(1)
Equating both sides and substituting x = 1 we get
y⁽¹⁾ = 0
Now assumed in the question are the values y⁽¹⁾ = 0 and y(2) = 1
We also know y(1) = √2
Putting them in equation (1) we get
2√2 y⁽³⁾ = 3! + 210 = 216
y⁽³⁾ = 54√2.

7. Let f(x)=\(\frac{e^x \times sin(x)}{x}\) 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
a) \(\frac{\pi n}{4}\)
b) \(\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\)
c) πn
d) \(\frac{\pi}{2n}\)
View Answer

Answer: b
Explanation: Expanding sin(x)/x into Taylor series we have
\(\frac{sin(x)}{x}=\frac{1}{1!}-\frac{x^2}{3!}+\frac{x^4}{5!}-…..\infty\)
Now Taking the n^th derivative of function using Leibniz rule we have
\((e^x(\frac{sin(x)}{x}))^{(n)}=c_0^ne^x(\frac{1}{1!}-\frac{x^2}{3!}+\frac{x^4}{5!}-…\infty)+c_1^ne^x(\frac{-2x}{3!}-\frac{4x^3}{5!}-…\infty)+….\)
Now substituting x=0 we have
\([(e^x(\frac{sin(x)}{x}))^{(n)}]_{x=0} = c_0^n(\frac{1}{1!})+c_2^n(\frac{-2!}{3!})+c_4n(\frac{4!}{5!})…\infty\)
\([(e^x(\frac{sin(x)}{x}))^{(n)}]_{x=0}=\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\)
Hence, Option \(\sum_{i=0}^{i<=n}\frac{(-1)^i c_{2i}^n}{2i+1}\) is the right answer.

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
a) \(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)
b) \(\sum_{i=0}^n \frac{1}{2i+1}\)
c) 1
d) Has no general form
View Answer

Answer: a
Explanation: The key here is to find a separate Taylor expansion for sinh(x) / x which is
\(\frac{sin h(x)}{x}=\frac{\frac{x}{1!}+\frac{x^3}{3!}+\frac{x^5}{5!}…\infty}{x}\)
\(\frac{sin h(x)}{x}=\frac{x}{1!}+\frac{x^2}{3!}+\frac{x^4}{5!}…\infty\)
Now consider \(((e^x)(\frac{sin h(x)}{x}))\) applying the Leibniz rule for n^th derivative we have
\((e^x(\frac{sin(x)}{x}))^{(n)}=c_0^ne^x(\frac{1}{1!}-\frac{x^2}{3!}+\frac{x^4}{5!}-…\infty)+c_1^ne^x(\frac{2x}{3!}+\frac{4x^3}{5!}…\infty)\)
\(+c_2^n(\frac{2}{3!} + \frac{12x^2}{5!}….\infty)+…\)
Now substituting x=0 yields
\((e^x(\frac{sin(x)}{x}))^{(n)}=\frac{c_0^n}{1} + \frac{c_2^n}{3} + \frac{c_4^n}{5}\)
\(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)

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