Question

Find the value of ${{y}_{2n}}$, If $y=(1-{{x}^{2}})\cos x$.

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 ${{y}_{2n}}$
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 $u=1-{{x}^{2}}$ and $v=\cos x$ .
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, $nth$ derivatives , ${{(n-1)}^{th}}$ derivatives, ${{(n-2)}^{nd}}$ derivative, and${{(n-3)}^{rd}}$derivatives,
So we get the derivatives as,
same for ${{v}_{1}}=-\sin x=\cos \left( \dfrac{\pi }{2}+x \right)$ , ${{v}_{2}}=-\sin \left( \dfrac{\pi }{2}+x \right)=\cos \left( \dfrac{\pi }{2}+\dfrac{\pi }{2}+x \right)=\cos \left( \dfrac{2\pi }{2}+x \right)$ ,
${{v}_{3}}=-\sin \left( \dfrac{2\pi }{2}+x \right)=\cos \left( \dfrac{\pi }{2}+\dfrac{2\pi }{2}+x \right)=\cos \left( \dfrac{3\pi }{2}+x \right)$ .
So at ${{v}_{n}}=\cos \left( \dfrac{n\pi }{2}+x \right)$ , ${{v}_{n-1}}=\cos \left( \dfrac{(n-1)\pi }{2}+x \right)$ ………(3)
So above we have made conversions don’t jumble in these conversions.
As we have find out the values of ${{u}_{1}},{{u}_{2}},{{u}_{3}}$ and ${{v}_{n}},{{v}_{n-1}},{{v}_{n-2}},{{v}_{n-3}}$ ,
So substituting (2) and (3) in (1), that is substituting in Leibnitz theorem,
So We get,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}((1-{{x}^{2}})\cos x)=(1-{{x}^{2}})\cos \left( \dfrac{n\pi }{2}+x \right)+{}^{n}{{c}_{1}}(-2x)\cos \left( \dfrac{(n-1)\pi }{2}+x \right)+{}^{n}{{c}_{2}}(-2)\cos \left( \dfrac{(n-2)\pi }{2}+x \right)$
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}((1-{{x}^{2}})\cos x)=(1-{{x}^{2}})\cos \left( \dfrac{n\pi }{2}+x \right)+n(-2x)\cos \left( \dfrac{(n-1)\pi }{2}+x \right)+\dfrac{n(n-1)}{2}(-2)\cos \left( \dfrac{(n-2)\pi }{2}+x \right)$
So simplifying in simple manner, we get,
$\dfrac{{{d}^{n}}}{d{{x}^{n}}}((1-{{x}^{2}})\cos x)=(1-{{x}^{2}})\cos \left( \dfrac{n\pi }{2}+x \right)-2nx\cos \left( \dfrac{(n-1)\pi }{2}+x \right)-n(n-1)\cos \left( \dfrac{(n-2)\pi }{2}+x \right)$
As it is given in question that $2n$ is used,
So We want to find ${{y}_{2n}}$ so substituting $2n$ in place of $n$ ,
So we get,
$\dfrac{{{d}^{2n}}}{d{{x}^{2n}}}((1-{{x}^{2}})\cos x)=(1-{{x}^{2}})\cos \left( \dfrac{2n\pi }{2}+x \right)-4nx\cos \left( \dfrac{(2n-1)\pi }{2}+x \right)-2n(2n-1)\cos \left( \dfrac{(2n-2)\pi }{2}+x \right)$
So the final answer is,
${{y}_{2n}}=(1-{{x}^{2}})\cos \left( \dfrac{2n\pi }{2}+x \right)-4nx\cos \left( \dfrac{(2n-1)\pi }{2}+x \right)-2n(2n-1)\cos \left( (n-1)\pi +x \right)$
So we can see in ${{y}_{2n}}$ ,
So if we take a term $\cos \left( \dfrac{2n\pi }{2}+x \right)=\cos \left( n\pi +x \right)$ ,
Of we take the value of $n$ as $2k$ i.e. even number the output we get as $\cos x$ ,
And if we take $n$ as $2k-1$ i.e. odd number the output we get as $-\cos x$ ,
So similarly if we take for term $\cos \left( \dfrac{(2n-1)\pi }{2}+x \right)$ ,
So if we take the value of $n$ as $2k$ i.e. even number the output we get as $\sin x$ ,
And for $n$ as $2k-1$ i.e. odd number the output we get as $-\sin x$,
So for the term $\cos \left( \dfrac{(2n-2)\pi }{2}+x \right)=\cos \left( (n-1)\pi +x \right)$ ,
So if we take the value of $n$ as $2k$ i.e. even number the output we get as $-\cos x$ ,
And for $n$ as $2k-1$ i.e. odd number the output we get as $\cos x$ ,
So in this way we get the trigonometric identities if $n$ is even or odd.

Note: Hence Be careful while solving the Leibnitz theorem. While solving confusion occurs. Be careful of using and converting sin and cos such as ${{v}_{1}}=-\sin x=\cos \left( \dfrac{\pi }{2}+x \right)$ . Also use proper $u$ and $v$ substitution. Substitute $u$ and $v$ in a proper manner, don't jumble yourself.