250+ TOP MCQs on Leibniz Rule and Answers

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

1. Let f(x) = ex sin(x2) ⁄ x Then the value of the fifth derivative at x = 0 is given by
a) 25
b) 21
c) 0
d) 5
Answer: b
Explanation: First expanding sin(x2) x into a Taylor series we have
sin(x2)=(frac{x^2}{1!}-frac{x^6}{3!}+frac{x^{10}}{5!}….infty)
(frac{sin(x^2)}{x}=frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}….infty)
Now applying the Leibniz rule up to the fifth derivative we have
(((e^x)(frac{sin(x^2)}{x})^{(5)} = c_0^5e^x(frac{x}{1!}-frac{x^5}{3!}+frac{x^9}{5!}…infty))
(+c_1^5e^x(frac{1}{1!}-frac{5x^4}{3!}+frac{9x^8}{5!}…infty)+…..+c_5^5e^x(frac{5!}{3!}+frac{(9.8.7.6.5)x^4}{5!}…infty))
Now substituting x=0 we get
(((e^x)(frac{sin(x^2)}{x}))^{(5)}=c_1^5(1)+frac{5!}{3!})
= 1 + 20 = 21.

2. Let f(x) = eex assuming all the nth derivatives at x =0 to be 1 the value of the (n + 1)th derivative can be written as
a) e – 1 + 2n
b) 0
c) 1
d) None
Answer: a
Explanation: Assume y = f(x)
Taking ln(x) on both sides The function has to be written in the form ln(y) = ex
Now computing the first derivative yields
y(1) = y * ex
Now applying the Leibniz rule up to nth derivative we have
y(n+1)=(c_0^ne^xy+c_1^ne^xy^{(1)}+….+c_n^ne^xy^{(n)})
We know that in the problem it is assumed that [y(1)=y(2)=…=y(n)=1]x=0
Now, substituting x=0 we get
y(n+1)=(c_0^ne+c_1^n+….+c_n^n)
From combinatorial results we know that 2n=(c_0^ne+c_1^n+….+c_n^n)
This gives us
y(n+1)=(e+(c_0^ne+c_1^n+….+c_n^n)-c_0^n)
y(n+1)=e-1+2n

3. Let f(x) = (sqrt{sin(x)}) and let yn denote the nth 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
Answer: a
Explanation: Assume y = f(x)
Rewriting the function as y2 = sin(x)
Now applying Leibniz rule up to the sixth derivative we have
(y2)(6) = c06 y(6) y + c16 y(5) y(1) + ………+ c66 y(6) y

(y2)(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) π2
c) 45
d) 4
Answer: a
Explanation: First convert the function sinh(x)⁄x into its Taylor series expansion
(frac{sinh(x)}{x}=frac{frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}….infty}{x})
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now pick up the whole function (((sin(x))(frac{sinh(x)}{x}))) and apply Leibniz rule up to the fourth derivative we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}=c_0^4sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}…..infty))
(-c_1^4cos(x)(frac{2x}{3!}+frac{4x^3}{5!}…..infty)+…..+c_4^4sin(x)(frac{4!}{5!}+frac{(7.6.5.4)x^3}{7!}…..infty))
Substituting x=0 we have
(((sin(x))(frac{sinh(x)}{x}))^{(4)}) = 0

5. The third derivative of f(x) = cos(x)sinh(x) ⁄ x at x = 0 is
a) 0
b) π32
c) (π)2
d) cos(1)sinh(1)
Answer: a
Explanation: Assume y = f(x)
Rewriting the part sinh(x)⁄x as infinite series we have
(frac{sinh(x)}{x}=frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)
Now the function f(x) becomes
y=(cos(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty))
Taking the third derivative of the above function using Leibniz rule we have
y(3)=(c_0^3sin(x)(frac{1}{1!}+frac{x^2}{3!}+frac{x^4}{5!}….infty)-c_1^3cos(x)(frac{2x}{3!}+frac{4x^3}{5!}….infty))
(-c_2^3sin(x)(frac{2}{3!}+frac{12x^2}{5!}….infty)+c_3^3cos(x)(frac{24x}{5!}…infty))
Now substituting x = 0 we have
y(3) = 0.

6. Let f(x) = (x2 + x + 1)sinh(x) the (1097)th derivative at x = 0 is
a) 1097
b) 1096
c) 0
d) 1202313
Answer: d
Explanation: Expanding sinh(x) into a taylor series we have
sinh(x)=(x+frac{x^3}{3!}+frac{x^5}{5!}…infty)
f(x)=(x2+x+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty))
On multiplication we get two series with odd exponents and one series with even exponent. The series with odd exponents are the only ones to contribute to the derivative at x=0
Hence it is enough to compute the derivative at for the following function
(x2+1)((x+frac{x^3}{3!}+frac{x^5}{5!}….infty)=frac{x}{1!}+x^3(frac{1}{3!}+1)+x^5(frac{1}{5!}+frac{1}{3!})….infty)
Taking the 1097thderivative of this function, we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!})+(1099times 1098…4times 3)x^2(frac{1}{(1097)!}+frac{1}{(1097)!})+…infty)
Substituting x=0 we have
f(1097)(x)=((1097)!(frac{1}{(1097)!}+frac{1}{(1095)!}))
=(1+1097*1096)=(1+1202312)=1202313

7. The 7th derivative of f(x) = (x3 + x2 + x + 1) sinh(x) at x = 0 is given by
a) 43
b) 7
c) 0
d) 34
Answer: a
Explanation: Expanding sinh(x) into a Taylor series we have
sinh(x)=(frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty)
Now rewriting the function we have
f(x)=(x3+x2+x+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
For the 7th derivative observe that, only the odd termed powers contribute to the derivative at x=0
Hence it is enough for us to find seventh derivative for
(x2+1)((frac{x}{1!}+frac{x^3}{3!}+frac{x^5}{5!}…infty))
(frac{x}{1!}+x^3(frac{1}{3!}+frac{1}{1!})+x^5(frac{1}{5!}+frac{1}{3!})+…infty)
Taking the 7th derivative of this function we have
f(7)(x)=((7!)(frac{1}{7!}+frac{1}{5!}) + (9*8…4*3)x^2(frac{1}{7!}+frac{1}{9!}))
Now substituting x=0 yields
f(7)(0)=((7!)(frac{1}{5!}+frac{1}{7!}))=(1+7*6)=43.

8. The (1071729)th derivative of f(x) = (x6 + x4 + x2) cosh(x) at x = 0 is given by
a) 0
b) 1071
c) 1729
d) ∞
Answer: a
Explanation: Expanding cosh(x) into a Taylor series we have
cosh(x)=(frac{1}{1!}+frac{x^2}{2!}+frac{x^4}{4!}…infty)
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) = (x2 + 1)tan-1 (x) at x = 0 is
a) 0
b) 1729
c) 1728
d) ∞
Answer: a
Explanation: Expanding the tan-1 (x) function into Taylor series we have
tan-1(x)=(frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty)
Rewrite the function as
f(x)=(x2+1)((frac{x}{1}-frac{x^3}{3}+frac{x^5}{5}-…infty))
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.

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