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,
or in Leibnitz's notation,
In different notation it can be written as,
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 $n=20$,
  & \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) \\
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 ${{v}_{1}} = \cos x=\sin \left( \dfrac {\pi }{2}+x \right)$.
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