1. Find the value of S=\(\sum_{n=1}^\infty \frac{(-1)^{n+1} \times (2n-1)^3}{(2n-1)!}\) using nth derivatives.
a) – 2 * sin(1)
b) 3 * sin(1)
c) 3 * cos(1)
d) – 3 * cos(1)
Explanation: We have to consider the function f(x) = sin(ex) in order to get the series in some way.
Expanding the given function into a Taylor series we have
f(x)=\(\frac{e^x}{1!}-\frac{e^{3x}}{3!}+\frac{e^{5x}}{5!}…\infty \)
Now observe that our series in question doesn’t have the exponential function, this gives us the hint that some derivative of this function has to be taken at x = 0
Observe that the term (2n – 1)3 has exponent equal to 3
Hence we have to take the third derivative of the function to get the required series
Now taking the third derivative yields
f(x)=\(\frac{e^x}{1!}-\frac{3^3e^{3x}}{3!}+\frac{5^5e^{5x}}{5!}…\infty \)
Now substituting x=0 we get
\(\frac{1^3}{1!}-\frac{3^3}{3!}+\frac{5^3}{5!}…\infty=\sum_{n=1}^\infty \frac{(-1)^{n+1} \times (2n-1)^3}{(2n-1)!} \)
To find the value of this series we need to take the third derivative of original function at the required point, this is as follows
f(3)(x) = -2exsin(ex)
Substituting x = 0 we get
f(3)(0) = -2sin(1).
2. Let f(x)=\(\frac{ln(1-x)}{e^x}\). Find the third derivative at x = 0.
a) 4
b) 1⁄3
c) Undefined
d) 1⁄4
Explanation: Again the key here is to expand the given function into appropriate Taylor series.
Rewriting the function as f(x) = e-x(ln(1 – x)) and then expanding into Taylor series we have
\(f(x)=(1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}…\infty) \times (\frac{x}{1}+\frac{x^2}{2}+\frac{x^3}{3}…\infty)\)
Now the question asks us to find the third derivative at x = 0. Thus, it is enough for us to find the coefficient of the x3 term in the infinite polynomial product above
The third degree terms can be grouped apart as follows
= x3⁄3 – x3⁄2 + x3⁄2
Hence the third derivative at x = 0 is simply the coefficient of the third degree term, which is
coefficient(x3⁄3) = 1⁄3.
3. Let f(x) = sin(x)/1+x2. Let y(n) denote the nth derivative of f(x) at x = 0 then the value of y(100) + 9900y(98) is
a) 0
b) -1
c) 100
d) 1729
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.
4. Let f(x) = ln(x)/x+1 and let y(n) denote the nth derivative of f(x) at x = 1 then the value of 2y(100) + 100y(99)
a) (99)!
b) (-99)!
c) (100)!
d) (-98)!
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 nth 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)!
5. Let f(x) = \(\sqrt{1-x^2}\) and let y(n) denote the nth 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
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.
6. Let f(x) = tan(x) and let y(n) denote the nth derivative of f(x) then the value of y(9998879879789776) is
a) 908090988
b) 0
c) 989
d) 1729
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 nnt 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.
7. 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
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.
8. 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
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.
9. Let f(x)=\(\frac{e^x \times sin(x)}{x}\) and let the nth 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) \(\frac{\pi}{2n}\)
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 nth 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.
10. Let f(x) = ex sinh(x) / x, let y(n) denote the nth 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
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 nth 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
\((e^x(\frac{sin(x)}{x}))^{(n)}=\frac{c_0^n}{1} + \frac{c_2^n}{3} + \frac{c_4^n}{5}\)
\(\sum_{i=0}^n \frac{c_{2i}^n}{2i+1}\)