Courses for Kids
Free study material
Offline Centres
Store Icon

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

Last updated date: 17th Apr 2024
Total views: 35.7k
Views today: 1.35k
35.7k+ views
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,
or in Leibnitz's notation,
In different notation it can be written as,
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) \\

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.