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Find the value of ${{y}_{20}}$, when $y={{x}^{3}}\sin x$, find ${{y}_{20}}$.

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Hint: In this question first we will select a term as $u$ and $v$. Differentiate it till when one of them gets zero by using Leibnitz theorem.

Complete step-by-step answer:

Here We have to find ${{y}_{20}}$
The product rule is a formula used to find the derivatives of products of two or more functions. It may be stated as,
${{(f.g)}^{'}}={{f}^{'}}.g+f.{{g}^{'}}$
or in Leibnitz's notation,
$\dfrac{d(u.v)}{dx}=\dfrac{du}{dx}.v+u.\dfrac{dv}{dx}$
In different notation it can be written as,
$d(uv)=udv+vdu$
The product rule can be considered a special case of the chain rule for several variables.
So the chain rule is,
$\dfrac{d(ab)}{dx}=\dfrac{\partial (ab)}{\partial a}\dfrac{da}{dx}+\dfrac{\partial (ab)}{\partial b}\dfrac{db}{dx}$
So we have to use the Leibnitz theorem,
So Leibnitz Theorem provides a useful formula for computing the ${{n}^{th}}$ derivative of a product of two functions. This theorem (Leibnitz theorem) is also called a theorem for successive differentiation.
This theorem is used for finding the ${{n}^{th}}$ derivative of a product. The Leibnitz formula expresses the derivative on ${{n}^{th}}$ order of the product of two functions.
If $y=u.v$ then $\dfrac{{{d}^{n}}}{d{{x}^{n}}}(u.v)=u{{v}_{n}}+{}^{n}{{c}_{1}}{{u}_{1}}{{v}_{n-1}}+{}^{n}{{c}_{2}}{{u}_{2}}{{v}_{n-2}}+.......+{}^{n}{{c}_{r}}{{u}_{r}}{{v}_{n-r}}+....+{{u}_{n}}{{v}_{{}}}$……(1)
Now Let us consider $u={{x}^{3}}$ and $v=\sin x$
Here now differentiating $u$ for first derivative ${{u}_{1}}$, then second derivative ${{u}_{2}}$ and then third derivative ${{u}_{3}}$.
So we get,
So ${{u}_{1}}=3{{x}^{2}}$, ${{u}_{2}}=6x$,${{u}_{3}}=6$,${{u}_{4}}=0$…..(2)
Also differentiating for $v$, For first , second,$nth$ derivatives , ${{(n-1)}^{th}}$ derivatives, ${{(n-2)}^{nd}}$ derivative, and ${{(n-3)}^{rd}}$derivatives,
So we get the derivatives as,
${{v}_{1}}=\cos x=\sin \left( \dfrac{\pi }{2}+x \right)$,${{v}_{2}}=\cos \left( \dfrac{\pi }{2}+x \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{2}+x \right)=\sin \left( \dfrac{2\pi }{2}+x \right)$,
${{v}_{3}}=\sin \left( \dfrac{3\pi }{2}+x \right)$
So at ${{v}_{n}}=\sin \left( \dfrac{n\pi }{2}+x \right)$,${{v}_{n-1}}=\sin \left( \dfrac{(n-1)\pi }{2}+x \right)$,${{v}_{n-2}}=\sin \left( \dfrac{(n-2)\pi }{2}+x \right)$,${{v}_{n-3}}=\sin \left( \dfrac{(n-3)\pi }{2}+x \right)$………(3)
So above I have made conversions don’t jumble in these conversions.
As we have find out the values of${{u}_{1}},{{u}_{2}},{{u}_{3}}$ and ${{v}_{n}},{{v}_{n-1}},{{v}_{n-2}},{{v}_{n-3}}$,
So substituting (2) and (3) in (1), that is substituting in Leibnitz theorem,
So we get,
$\Rightarrow$ $\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{3}})\sin x)={{x}^{3}}\sin \left( \dfrac{n\pi }{2}+x \right)+{}^{n}{{c}_{1}}3{{x}^{2}}\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+{}^{n}{{c}_{2}}6x\sin \left( \dfrac{(n-2)\pi }{2}+x \right)+{}^{n}{{c}_{3}}6\sin \left( \dfrac{(n-3)\pi }{2}+x \right)$
$\Rightarrow$ \[\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{3}})\sin x)={{x}^{3}}\sin \left( \dfrac{n\pi }{2}+x \right)+3n{{x}^{2}}\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+\dfrac{n(n-1)}{2}6x\sin \left( \dfrac{(n-2)\pi }{2}+x \right)+\dfrac{n(n-1)(n-2)}{6}6\sin \left( \dfrac{(n-3)\pi }{2}+x \right)\]
So simplifying in simple manner we get,
$\Rightarrow$ $\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{3}})\sin x)={{x}^{3}}\sin \left( \dfrac{n\pi }{2}+x \right)+3n{{x}^{2}}\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+3n(n-1)x\sin \left( \dfrac{(n-2)\pi }{2}+x \right)+n(n-1)(n-2)\sin \left( \dfrac{(n-3)\pi }{2}+x \right)$
Here$n=20$so substituting $n$ as$20$, As it is given in question as${{y}_{20}}$, so here $n=20$,
$\begin{align}
  & \dfrac{{{d}^{20}}}{d{{x}^{20}}}(({{x}^{3}})\sin x)={{x}^{3}}\sin \left( \dfrac{20\pi }{2}+x \right)+60{{x}^{2}}\sin \left( \dfrac{(20-1)\pi }{2}+x \right)+60(20-1)x\sin \left( \dfrac{(20-2)\pi }{2}+x \right)+20(20-1)(20-2)\sin \left( \dfrac{(20-3)\pi }{2}+x \right) \\
 & \dfrac{{{d}^{20}}}{d{{x}^{20}}}(({{x}^{3}})\sin x)={{x}^{3}}\sin \left( \dfrac{20\pi }{2}+x \right)+60{{x}^{2}}\sin \left( \dfrac{19\pi }{2}+x \right)+1140x\sin \left( \dfrac{18\pi }{2}+x \right)+6840\sin \left( \dfrac{17\pi }{2}+x \right) \\
\end{align}$
Hence we get solution as,
Hence ${{y}_{20}}={{x}^{3}}\sin \left( \dfrac{20\pi }{2}+x \right)+60{{x}^{2}}\sin \left( \dfrac{19\pi }{2}+x \right)+1140x\sin \left( \dfrac{18\pi }{2}+x \right)+6840\sin \left( \dfrac{17\pi }{2}+x \right)$
So simplifying we get,
\[\sin \left( \dfrac{20\pi }{2}+x \right)=\sin x,\sin \left( \dfrac{19\pi }{2}+x \right)=-\cos x,\sin \left( \dfrac{18\pi }{2}+x \right)=-\sin x,\sin \left( \dfrac{17\pi }{2}+x \right)=\cos x\]
So substituting we get the final answer as,

${{y}_{20}}={{x}^{3}}\sin x-60{{x}^{2}}\cos x-1140x\sin x+6840\cos x$

Note: Understand the question first: what is ${{y}_{20}}$. Be careful while solving the Leibnitz theorem. Substitute the proper values of $u$ and $v$. While solving or converting you must take care of signs and angles. So take utmost care while conversion of$\sin $into$\cos $or vice-versa such as ${{v}_{1}} = \cos x=\sin \left( \dfrac {\pi }{2}+x \right)$.