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Find the value of ${{y}_{n}}$, when $y={{e}^{x}}(3{{x}^{2}}-4)$, Find ${{y}_{n}}$.

Last updated date: 13th Jul 2024
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Hint: In this question first we will Select a term as $u$ and $v$. Differentiate it till you get zero by using Leibnitz theorem.

Complete step-by-step answer:
So to find ${{y}_{n}}$ means to find $\dfrac{{{d}^{n}}y}{d{{x}^{n}}}$,
So for nth derivative we know that we should use Leibnitz theorem,
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)
The above theorem is Leibnitz theorem,
So Now Let us consider $u=3{{x}^{2}}-4$ and $v={{e}^{x}}$
Here now differentiating$u$for first derivative${{u}_{1}}$,then second derivative${{u}_{2}}$and then third derivative${{u}_{3}}$.
So we get the derivatives as,
So we get ${{u}_{1}},{{u}_{2}},{{u}_{3}}$,
So ${{u}_{1}}=6x$, ${{u}_{2}}=6$,${{u}_{3}}=0$…..(2)
Also differentiating for$v$, For first , second,${{n}^{th}}$derivatives , ${{(n-1)}^{th}}$derivatives, ${{(n-2)}^{nd}}$derivative,
So we get the derivatives as,
same for ${{v}_{1}}={{e}^{x}}$,${{v}_{2}}={{e}^{x}}$, ${{v}_{3}}={{e}^{x}}$ So${{v}_{n}}={{e}^{x}},{{v}_{n-1}}={{e}^{x}},{{v}_{n-2}}={{e}^{x}}$ ………(3)
Now we have to substitute (2) and (3) in (1), that is substituting in Leibnitz theorem,
We get,
$\Rightarrow$ $\dfrac{{{d}^{n}}}{d{{x}^{n}}}((3{{x}^{2}}-4){{e}^{x}})=(3{{x}^{2}}-4){{e}^{x}}+{}^{n}{{c}_{1}}6x{{e}^{x}}+{}^{n}{{c}_{2}}6{{e}^{x}}$
$\Rightarrow$ $\dfrac{{{d}^{n}}}{d{{x}^{n}}}((3{{x}^{2}}-4){{e}^{x}})=(3{{x}^{2}}-4){{e}^{x}}+n6x{{e}^{x}}+\dfrac{n(n-1)}{2}6{{e}^{x}}$………………(we know${}^{n}{{c}_{1}}=n$and ${}^{n}{{c}_{2}}=\dfrac{n(n-1)}{2}$)
So simplifying in simple manner we get,
  & \dfrac{{{d}^{n}}}{d{{x}^{n}}}((3{{x}^{2}}-4){{e}^{x}})=(3{{x}^{2}}-4){{e}^{x}}+6nx{{e}^{x}}+3n(n-1){{e}^{x}} \\
 & \dfrac{{{d}^{n}}}{d{{x}^{n}}}((3{{x}^{2}}-4){{e}^{x}})={{e}^{x}}((3{{x}^{2}}-4)+6nx+3n(n-1)) \\
As we want to find for${{n}^{th}}$ derivative, So we get the final answer as,
Hence ${{y}_{n}}={{e}^{x}}((3{{x}^{2}}-4)+6nx+3n(n-1))$

Note: Be careful using Leibnitz theorem. Use proper substitution of $u$ and $v$. Don’t be confused while applying the $u$ and $v$. While solving confusion occurs. Use the differentiation in the correct manner. Be thorough with ${}^{n}{{c}_{1}}=n$and more. Use proper substitution of$u$and $v$. Don’t jumble between ${{u}_{1}}=6x$,${{u}_{2}}=6$etc.