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

Last updated date: 29th Mar 2023

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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,

${{(f.g)}^{'}}={{f}^{'}}.g+f.{{g}^{'}}$

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)

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,

$\begin{align}

& \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)) \\

\end{align}$

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.

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,

${{(f.g)}^{'}}={{f}^{'}}.g+f.{{g}^{'}}$

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)

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,

$\begin{align}

& \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)) \\

\end{align}$

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.

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