Question

Y=xln(x-1/x+1)
Find nth derivatives

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\green{y=x\,\ln\bigg(\dfrac{x-1}{x+1}\bigg)}}$

$\tt{\implies\,y=x\,\{\ln(x-1)-\ln(x+1)\}}$

$\tt{\implies\,y=x\,\ln(x-1)-x\,\ln(x+1)\,\,\,\,\,\,\,\,...(1)}$

Now, let us find $n^{th}$ derivative of $\sf{ln(x+a)}$

Consider the function $\sf{f(x)=ln(x+a)}$, we have,

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

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

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

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

So,

$\sf{\implies\,\blue{f^{(n)}(x)=\dfrac{(-1)^{n-1}\cdot(n-1)!}{(x+a)^{n}}}\,\,\,\,\,\,...(2)}$

Now, from (1), let us take

$\tt{g(x)=x\,\ln(x-1)\,\,\,\,\,\,and\,\,\,\,\,\,h(x)=x\,\ln(x+1)}$

Now, to find $\sf{y_{n}}$, we must find $\sf{g^{(n)}(x)\,\,\,\,and\,\,\,\,h^{(n)}(x)}$

$\tt{\bold{\purple{Finding\,\,\,n^{th}\,\,\, derivative\,\,\,of\,\,\,g(x)\,:}}}$

$\sf{g(x)=x\,\ln(x-1)}$

From Leibnitz's Theorem,

$\sf{If\,\,y=uv,\,then}\\\sf{y_{n}=u_{n}\cdot\,v+\,^{n}C_{1}\cdot\,u_{n-1}\cdot\,v_{1}+\,^{n}C_{2}\cdot\,u_{n-2}\cdot\,v_{2}+\,^{n}C_{3}\cdot\,u_{n-3}\cdot\,v_{3}+....+u\cdot\,v_{n+1}}$

On g(x), taking  $\sf{u=\ln(x-1)\,\,\,\,and\,\,\,\,v=x}$

$\sf{g^{(n)}(x)=x\cdot\dfrac{d^{n}}{dx^{n}}\{\ln(x-1)\}+\,^{n}C_{1}\cdot\dfrac{d^{n-1}}{dx^{n-1}}\{\ln(x-1)\}\cdot\dfrac{d}{dx}(x)+\,^{n}C_{2}\cdot\dfrac{d^{n-2}}{dx^{n-2}}\{\ln(x-1)\}\cdot\dfrac{d}{dx}(1)+..\,}$

$\sf{\implies\,g^{(n)}(x)=x\cdot\dfrac{d^{n}}{dx^{n}}\{\ln(x-1)\}+\,^{n}C_{1}\cdot\dfrac{d^{n-1}}{dx^{n-1}}\{\ln(x-1)\}\cdot\dfrac{d}{dx}(x)+0}$

From (2), we get,

$\sf{\implies\,g^{(n)}(x)=x\cdot\dfrac{(-1)^{n-1}\cdot(n-1)!}{(x-1)^{n}}+n\cdot\,\dfrac{(-1)^{n-2}\cdot(n-2)!}{(x-1)^{n-1}}\cdot1}$

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

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

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

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

$\sf{\pink{\implies\,g^{(n)}(x)=\dfrac{(-1)^{n-1}\cdot(n-x)\cdot(n-2)!}{(x-1)^{n}}}}$

Similarly, for h(x), we have,

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

So,

$\sf{y_{n}=g^{(n)}(x)-h^{(n)}(x)}$

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

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

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