Question

Find the nth derivative of y=1/x^2-6x+8

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=\dfrac{1}{x^2-6x+8}}$

$\tt{\implies\,y=\dfrac{1}{x^2-4x-2x+8}}$

$\tt{\implies\,y=\dfrac{1}{x(x-4)-2(x-4)}}$

$\tt{\implies\,y=\dfrac{1}{(x-2)(x-4)}}$

$\sf{\implies\,y=\dfrac{2}{2(x-2)(x-4)}}$

$\sf{\implies\,y=\dfrac{(x-2)-(x-4)}{2(x-2)(x-4)}}$

$\sf{\implies\,y=\dfrac{(x-2)}{2(x-2)(x-4)}-\dfrac{(x-4)}{2(x-2)(x-4)}}$

$\sf{\implies\,y=\dfrac{1}{2(x-4)}-\dfrac{1}{2(x-2)}}$

Consider the generalized form

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

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

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

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

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

So,

$\sf{\implies\,y_{n}=\dfrac{(-1)^n\cdot\,n!}{2(x-4)^{n+1}}-\dfrac{(-1)^n\cdot\,n!}{2(x-2)^{n+1}}}$