Question

Find the value of $\dfrac{{{d}^{24}}y}{d{{x}^{24}}}$, If $y={{e}^{x}}({{x}^{2}}-1)$ .

Answer 1

Hint: The differentiation to be found can be found using the Leibnitz rule for differentiation. The Leibnitz rule for finding successive differentiation can be stated as the derivative of ${{n}^{th}}$ order is given by the following rule:
If $y=u.v$ then $\frac{{{d}^{n}}}{d{{x}^{n}}}(u.v)=u{{v}_{n}}+{}^{n}{{c}_{1}}{{u}_{1}}{{v}_{n-1}}+{}^{n}{{c}_{2}}{{u}_{2}}{{v}_{n-2}}+.......+{}^{n}{{c}_{r}}{{u}_{r}}{{v}_{n-r}}+....+{{u}_{n}}{{v}_{{}}}$.

Complete step-by-step solution -
Here we have to find $\dfrac{{{d}^{24}}y}{d{{x}^{24}}}$.
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,
{\displaystyle {\dfrac {d}{dx}}(u\cdot v)={\dfrac {du}{dx}}\cdot v+u\cdot {\dfrac {dv}{dx}}.}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 $u={{x}^{2}}-1$ and $v={{e}^{x}}$ ,
So now we have to differentiate,
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 ${{u}_{1}}=2x$ , ${{u}_{2}}=2$ , ${{u}_{3}}=0$ …..(2)
Also differentiating for $v$ , For first , second, ${{n}^{th}}$ derivatives and ${{(n-1)}^{th}}$ derivatives,
So we get the derivatives as,
${{v}_{1}}={{e}^{x}}$ , ${{v}_{2}}={{e}^{x}}$ , ${{v}_{3}}={{e}^{x}}$ So ${{v}_{n}}={{e}^{x}}$ ………(3)
As we have find out the values of ${{u}_{1}},{{u}_{2}},{{u}_{3}}$ and ${{v}_{n}},{{v}_{n-1}},{{v}_{n-2}}$ ,
Now we have to substitute (2) and (3) in (1) that is substituting in Leibnitz theorem,
So after substituting we get the equation as,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+{}^{n}{{c}_{1}}2x{{e}^{x}}+{}^{n}{{c}_{2}}2{{e}^{x}}$
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+n2x{{e}^{x}}+\dfrac{n(n-1)}{2}2{{e}^{x}}$ ………………(we know ${}^{n}{{c}_{1}}=n$ and ${}^{n}{{c}_{2}}=\dfrac{n(n-1)}{2}$)
So simplifying in simple manner we get,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+2nx{{e}^{x}}+n(n-1){{e}^{x}}$
Now substituting $n=24$ , as it is given in question,
So substituting $n=24$ we get the following,
$\begin{aligned} & \dfrac{{{d}^{24}}}{d{{x}^{24}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+48x{{e}^{x}}+24(24-1){{e}^{x}} \\\ & \dfrac{{{d}^{24}}}{d{{x}^{24}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+48x{{e}^{x}}+24(23){{e}^{x}} \\\ & \dfrac{{{d}^{24}}}{d{{x}^{24}}}(({{x}^{2}}-1){{e}^{x}})=({{x}^{2}}-1){{e}^{x}}+48x{{e}^{x}}+552{{e}^{x}} \\\ & \dfrac{{{d}^{24}}}{d{{x}^{24}}}(({{x}^{2}}-1){{e}^{x}})={{e}^{x}}(({{x}^{2}}-1)+48x+552) \\\ & \\\ \end{aligned}$
So we get final answer as,
Hence $\dfrac{{{d}^{24}}}{d{{x}^{24}}}({{x}^{2}}-1){{e}^{x}}={{e}^{x}}(({{x}^{2}}-1)+48x+552)$

Note: Be careful while solving the Leibnitz theorem not a single value should be missed. Be careful while substituting the value of $n$ . Also take care while substituting $u$ and $v$ . Don’t make mistakes while differentiating $u$ and $v$ . Be thorough with ${}^{n}{{c}_{1}}=n$ and more. Don’t jumble yourself while simplifying.