Question

Find the nth derivative of y= 1/(1+x+x^2+x^3).​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=\dfrac{1}{1+x+x^2+x^3}}$

$\sf{\implies\,y=\dfrac{1}{(1+x)+x^2(1+x)}}$

$\sf{\implies\,y=\dfrac{1}{(x+1)(x^2+1)}}$

$\sf{\implies\,y=\dfrac{1}{(x+1)(x+i)(x-i)}}$

Decompose into partial fractions

$\sf{\implies\,y=\dfrac{1}{(x+1)(-1+i)(-1-i)}+\dfrac{1}{(1-i)(x+i)(-i-i)}+\dfrac{1}{(1+i)(i+i)(x-i)}}$

$\sf{\implies\,y=\dfrac{1}{(x+1)(1-i)(1+i)}+\dfrac{1}{(i-1)(x+i)(i+i)}+\dfrac{1}{(1+i)(i+i)(x-i)}}$

$\sf{\implies\,y=\dfrac{1}{(x+1)(1-i^2)}+\dfrac{1}{(i-1)(x+i)(2i)}+\dfrac{1}{(1+i)(2i)(x-i)}}$

$\sf{\implies\,y=\dfrac{1}{2(x+1)}+\dfrac{1}{(2i^2-2i)(x+i)}+\dfrac{1}{(2i+2i^2)(x-i)}}$

$\sf{\implies\,y=\dfrac{1}{2(x+1)}+\dfrac{1}{(-2-2i)(x+i)}+\dfrac{1}{(2i-2)(x-i)}}$

Now, the nth derivative of y will be the sum of the nth derivative of the above factors

So,

Consider the generalized form

$\sf{f(x)=\dfrac{1}{x+a}}$

$\sf{\implies\,f^{\prime}(x)=\dfrac{-1}{(x+a)^2}}$

$\sf{\implies\,f^{\prime\prime}(x)=\dfrac{(-1)(-2)}{(x+a)^3}}$

$\sf{\implies\,f^{\prime\prime\prime}(x)=\dfrac{(-1)(-2)(-3)}{(x+a)^4}}$

So,

$\sf{f^{n}(x)=\dfrac{(-1)^n\cdot\,n!}{(x+a)^{n+1}}}$

$\sf{\implies\,y_{n}=\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{2(x+1)}\right\}+\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{(-2-2i)(x+i)}\right\}+\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{(2i-2)(x-i)}\right\}}$

$\sf{\implies\,y_{n}=\dfrac{1}{2}\cdot\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{x+1}\right\}+\dfrac{1}{(-2-2i)}\cdot\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{x+i}\right\}+\dfrac{1}{(2i-2)}\cdot\dfrac{d^{n}}{dx^{n}}\left\{\dfrac{1}{x-i}\right\}}$

$\sf{\implies\,y_{n}=\dfrac{1}{2}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x+1)^{n+1}}+\dfrac{1}{(-2-2i)}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x+i)^{n+1}}+\dfrac{1}{(2i-2)}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x-i)^{n+1}}}$

$\sf{\implies\,y_{n}=\dfrac{1}{2}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x+1)^{n+1}}+(-1)^{n}\cdot\,n!\left\{\dfrac{1}{(-2-2i)}\cdot\dfrac{1}{(x+i)^{n+1}}+\dfrac{1}{(2i-2)}\cdot\dfrac{1}{(x-i)^{n+1}}\right\}}$

To remove the imaginary terms, put x= r cos(θ) and 1= r sin(θ)

where, $\sf{r=\sqrt{x^2+1}}\,\,\,\,\,and\,\,\,\,\,\theta=cot^{-1}(x)$

Let us solve only the terms in braces

$\sf{\left\{\dfrac{1}{-2\left(\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}i\right)}\cdot\dfrac{1}{(r\,cos(\theta)+r\,i\,sin(\theta))^{n+1}}+\dfrac{1}{-2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{2}}i\right)}\cdot\dfrac{1}{(r\,cos(\theta)-r\,i\,sin(\theta))^{n+1}}\right\}}$

$\sf{=\left\{\dfrac{1}{-2\,e^{\frac{i\,\pi}{4}}}\cdot\dfrac{1}{r^{n+1}(cos(\theta)+i\,sin(\theta))^{n+1}}+\dfrac{1}{-2e^{-\frac{i\,\pi}{4}}}\cdot\dfrac{1}{r^{n+1}(cos(\theta)-i\,sin(\theta))^{n+1}}\right\}}$

$\sf{=\left\{\dfrac{1}{-2\,e^{\frac{i\,\pi}{4}}}\cdot\dfrac{1}{r^{n+1}\cdot\,e^{i(n+1)\theta}}+\dfrac{1}{-2e^{-\frac{i\,\pi}{4}}}\cdot\dfrac{1}{r^{n+1}\cdot\,e^{-i(n+1)\theta}}\right\}}$

$\sf{=-\dfrac{1}{2\,r^{n+1}}\left\{\dfrac{1}{e^{\frac{i\,\pi}{4}}}\cdot\dfrac{1}{e^{i(n+1)\theta}}+\dfrac{1}{e^{-\frac{i\,\pi}{4}}}\cdot\dfrac{1}{e^{-i(n+1)\theta}}\right\}}$

$\sf{=-\dfrac{1}{2\,r^{n+1}}\left\{\dfrac{1}{e^{i\{(n+1)\theta+\frac{\pi}{4}\}}}+\dfrac{1}{e^{-i\{(n+1)\theta+\frac{\pi}{4}\}}}\right\}}$

$\sf{=-\dfrac{1}{2\,r^{n+1}}\left\{e^{-i\{(n+1)\theta+\frac{\pi}{4}\}}+e^{i\{(n+1)\theta+\frac{\pi}{4}\}}\right\}}$

$\sf{=-\dfrac{1}{r^{n+1}}\left\{\dfrac{e^{i\{(n+1)\theta+\frac{\pi}{4}\}}+e^{-i\{(n+1)\theta+\frac{\pi}{4}\}}}{2}\right\}}$

$\sf{=-\dfrac{1}{r^{n+1}}\cdot\,cos\left\{(n+1)\theta+\dfrac{\pi}{4}\right\}}$

$\sf{=-\dfrac{1}{\left(\sqrt{x^2+1}\right)^{n+1}}\cdot\,cos\left\{(n+1)\,cot^{-1}(x)+\dfrac{\pi}{4}\right\}}$

Now,

$\sf{\implies\,y_{n}=\dfrac{1}{2}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x+1)^{n+1}}+(-1)^{n}\cdot\,n!\left[-\dfrac{1}{\left(\sqrt{x^2+1}\right)^{n+1}}\cdot\,cos\left\{(n+1)\,cot^{-1}(x)+\dfrac{\pi}{4}\right\}\right]}$$\sf{\implies\,y_{n}=\dfrac{1}{2}\cdot\dfrac{(-1)^{n}\cdot\,n!}{(x+1)^{n+1}}+\dfrac{(-1)^{n+1}\cdot\,n!}{\left(\sqrt{x^2+1}\right)^{n+1}}\cdot\,cos\left\{(n+1)\,cot^{-1}(x)+\dfrac{\pi}{4}\right\}}$