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

Answer

Verified

362.1k+ views

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.

Last updated date: 29th Sep 2023

•

Total views: 362.1k

•

Views today: 9.62k

Recently Updated Pages

What do you mean by public facilities

Difference between hardware and software

Disadvantages of Advertising

10 Advantages and Disadvantages of Plastic

What do you mean by Endemic Species

What is the Botanical Name of Dog , Cat , Turmeric , Mushroom , Palm

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Why are resources distributed unequally over the e class 7 social science CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Briefly mention the contribution of TH Morgan in g class 12 biology CBSE

What is the past tense of read class 10 english CBSE