Question

find the nth derivative of the function 6/(x^2-4) (x-1)​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\rm{y=\dfrac{6}{\left({x}^{2}-4\right)\left(x-1\right)}}$

$\rm{\implies\,y=\dfrac{6}{\left(x-2\right)\left(x+2\right)\left(x-1\right)}}$

On decomposing into partial fractions,

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

We know,

$\boxed{\sf{\dfrac{{d}^{n}}{d{x}^{n}}\left(\dfrac{1}{x+a}\right)=\dfrac{{(-1)}^{n}\cdot\,n!}{\left(x+a\right)^{n+1}}}}$

So,

$\rm{\implies\,{y}_{n}=\dfrac{\left(-1\right)^{n}\cdot2\cdot{n}!}{\left(x-2\right)^{n+1}}+\dfrac{\left(-1\right)^{n}\cdot{n}!}{2\left(x+2\right)^{n+1}}-\dfrac{\left(-1\right)^{n}\cdot2\cdot{n}!}{\left(x-1\right)^{n+1}}}$