Question

Find the value of $\dfrac{{{d}^{25}}y}{d{{x}^{25}}}$, If $y={{x}^{2}}\sin x$.

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 $\dfrac{{{d}^{25}}y}{d{{x}^{25}}}$ .
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 $u={{x}^{2}}$ and $v=\sin x$ .
Here now differentiating $u$ for first derivative ${{u}_{1}}$ ,then second derivative ${{u}_{2}}$ and then third derivative ${{u}_{3}}$ .
So we get,
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,
same for ${{v}_{1}}=\cos x=\sin \left( \dfrac{\pi }{2}+x \right)$ , ${{v}_{2}}=\cos \left( \dfrac{\pi }{2}+x \right)=\sin \left( \dfrac{\pi }{2}+\dfrac{\pi }{2}+x \right)=\sin \left( \dfrac{2\pi }{2}+x \right)$ ,
So above I have made conversions don’t jumble in this conversions.
So at ${{v}_{n}}=\sin \left( \dfrac{n\pi }{2}+x \right)$ , ${{v}_{n-1}}=\sin \left( \dfrac{(n-1)\pi }{2}+x \right)$ , ${{v}_{n-2}}=\sin \left( \dfrac{(n-2)\pi }{2}+x \right)$ ………(3)
As we have find out the values of ${{u}_{1}},{{u}_{2}},{{u}_{3}}$ and ${{v}_{n}},{{v}_{n-1}},{{v}_{n-2}}$ ,
So substituting (2) and (3) in (1), that is substituting in formula of Leibnitz theorem,
So, We get,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{n\pi }{2}+x \right)+{}^{n}{{c}_{1}}2x\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+{}^{n}{{c}_{2}}2\sin \left( \dfrac{(n-2)\pi }{2}+x \right)$
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{n\pi }{2}+x \right)+2nx\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+\dfrac{n(n-1)}{2}2\sin \left( \dfrac{(n-2)\pi }{2}+x \right)$
So simplifying in simple manner, we get,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{n\pi }{2}+x \right)+2nx\sin \left( \dfrac{(n-1)\pi }{2}+x \right)+n(n-1)\sin \left( \dfrac{(n-2)\pi }{2}+x \right)$
In question it is mention $\dfrac{{{d}^{25}}y}{d{{x}^{25}}}$ that means we have to find it for $n=25$ ,
Here $n=25$ so substituting $n$ as $25$ ,

& \dfrac{{{d}^{25}}}{d{{x}^{25}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{25\pi }{2}+x \right)+50x\sin \left( \dfrac{(25-1)\pi }{2}+x \right)+25(25-1)\sin \left( \dfrac{(25-2)\pi }{2}+x \right) \\\ & \dfrac{{{d}^{25}}}{d{{x}^{25}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{25\pi }{2}+x \right)+50x\sin \left( \dfrac{24\pi }{2}+x \right)+600\sin \left( \dfrac{23\pi }{2}+x \right) \\\ \end{aligned}$$ Hence we get the answer as, $\dfrac{{{d}^{25}}}{d{{x}^{25}}}(({{x}^{2}})\sin x)={{x}^{2}}\sin \left( \dfrac{25\pi }{2}+x \right)+50x\sin \left( \dfrac{24\pi }{2}+x \right)+600\sin \left( \dfrac{23\pi }{2}+x \right)$ So again simplifying the answer we get, $\sin \left( \dfrac{25\pi }{2}+x \right)=\cos x,\sin \left( \dfrac{24\pi }{2}+x \right)=\sin x,\sin \left( \dfrac{23\pi }{2}+x \right)=-\cos x$ So the final answer becomes, $\dfrac{{{d}^{25}}}{d{{x}^{25}}}(({{x}^{2}})\sin x)={{x}^{2}}\cos x+50x\sin x-600\cos x$ Note: Be careful while solving the Leibnitz theorem. While solving confusion occurs. Use the differentiation in the correct manner. Take care of the signs. Also take care while substituting $u$ and $v$. Don’t make mistakes while differentiating $u$ and $v$ . Be clear with the conversion of $\sin $ to $\cos $ such as ${{v}_{1}}=\cos x=\sin \left( \dfrac{\pi }{2}+x \right)$ See here how it is done. Don’t confuse yourself.