Question

By using a Mac Laurin series, compute the $9th$ derivative of $\arctan \left( {\dfrac{{{x^3}}}{2}} \right)$ at $x = 0$ .

Answer 1

Hint : The infinite sum of terms that are represented in terms of the function’s derivatives at a single point are said to be the taylor series of a function. The Taylor series is introduced by Brook Taylor in $1715$ . A taylor series is also called the Maclaurin series, if the derivatives are considered from where the point is zero.
Formula used: The MacLaurin series of $\arctan \left( x \right)$ is,
$\arctan \left( x \right) = x - \dfrac{{{x^3}}}{3} + \dfrac{{{x^5}}}{5} - \dfrac{{{x^7}}}{7} + ....... +$

Complete step by step solution:
In this problem, we have to compute the $9th$ derivative of $\arctan \left( {\dfrac{{{x^3}}}{2}} \right)$ at $x = 0$ by using Mac Laurin series.
On comparing the variable given in the question with the variable of the Mac Laurin series, we have to replace the variable x with $\dfrac{{{x^3}}}{2}$ .Then, the series becomes,
$\Rightarrow \arctan \left( {\dfrac{{{x^3}}}{2}} \right) = \dfrac{{{x^3}}}{2} - \dfrac{{{{\left( {\dfrac{{{x^3}}}{2}} \right)}^3}}}{3} + \dfrac{{{{\left( {\dfrac{{{x^3}}}{2}} \right)}^5}}}{5} - \dfrac{{{{\left( {\dfrac{{{x^3}}}{2}} \right)}^7}}}{7} + ....... +$
On further solving the series, we get,
$\Rightarrow \arctan \left( {\dfrac{{{x^3}}}{2}} \right) = \dfrac{{{x^3}}}{2} - \dfrac{{\left( {\dfrac{{{x^9}}}{8}} \right)}}{3} + \dfrac{{{{\left( {\dfrac{{{x^{15}}}}{{32}}} \right)}^{}}}}{5} - ....... +$
Now, the series will look like this,
$\arctan \left( {\dfrac{{{x^3}}}{2}} \right) = \dfrac{{{x^3}}}{2} - \dfrac{{{x^9}}}{{24}} + \dfrac{{{x^{15}}}}{{160}} - ....... +$
We know that in the Mac Laurin series, the n-th derivative of the f(x) evaluated in 0 is the coefficient of each term of n degree. Then, it becomes,
$\dfrac{{{f^{\left( 9 \right)}}\left( 0 \right)}}{{9!}} = - \dfrac{1}{{24}}$
By taking $9!$ on other side, we get,
${f^{9th}}\left( 0 \right) = - \dfrac{1}{{24}} \times 9!$
On further solving, we get,
$\Rightarrow {f^{\left( 9 \right)}}\left( 0 \right) = - \dfrac{{9!}}{{24}} \\\ \Rightarrow {f^{\left( 9 \right)}}\left( 0 \right) = - \dfrac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{24}} \\\ \Rightarrow {f^{\left( 9 \right)}}\left( 0 \right) = - \dfrac{{362880}}{{24}} = - 15120 \;$
Hence, the ${9^{th}}$ derivative of $\arctan \left( {\dfrac{{{x^3}}}{2}} \right)$ at $x = 0$ by using Mac Laurin series is $- 15120$ .
So, the correct answer is “ $- 15120$ ”.

Note : Mac Laurin series plays an important role in finding the derivative of $\arctan \left( {\dfrac{{{x^3}}}{2}} \right)$ at $x = 0$ . In this problem, firstly we have substituted the value $\dfrac{{{x^3}}}{2}$ on the place of x in the Mac Laurin series and we have to find the $9th$ derivative, so, after solving it, the ${9^{th}}$ derivative of $\arctan \left( {\dfrac{{{x^3}}}{2}} \right)$ at $x = 0$ is the coefficient of the each term of $9$ degree.