Answer
Verified
478.5k+ 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.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE