This set of 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
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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
view answer
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
view answer
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) π⁄2
c) 45
d) 4
view answer
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) π⁄32
c) (π)2
d) cos(1)sinh(1)
view answer
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
view answer
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)=\((1097)!(\frac{1}{(1097)!}+\frac{1}{(1095)!})+(1099\times 1098…4\times 3)x^2(\frac{1}{(1097)!}+\frac{1}{(1097)!})+…\infty\)
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
view answer
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) ∞
view answer
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) ∞
view answer
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.
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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