Question

Find the nth derivative of x/(x-1)(2x+1).

Answer 1

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Answer 2

Answer:

The nth derivative of $\frac{x}{(x-1)(2x+1)}$ is  $\frac{1}{3}[\frac{(-1)^n*n!}{(x-1)^{n+1}} +\frac{(-1)^n*2^n*n!}{(2x+1)^{n+1}} ]$

Step-by-step explanation:

Let y = $\frac{x}{(x-1)(2x+1)}$

        = $\frac{1}{3}[\frac{1}{x-1} +\frac{1}{2x+1} ]$

Differentiating with respect to x

=> $\frac{dy}{dx}$ =  $\frac{1}{3}[\frac{-1}{(x-1)^2} +\frac{-1*2}{(2x+1)^2} ]$                                  (as $\frac{d}{dx} (x^{n} ) = nx^{n-1}$)

Differentiating again with respect to x

=> $\frac{d^2y}{dx^2}$ =  $\frac{1}{3}[\frac{-1*(-2)}{(x-1)^3} +\frac{(-1*2)(-2*2)}{(2x+1)^3} ]$

Differentiating again with respect to x

=> $\frac{d^3y}{dx^3}$ =  $\frac{1}{3}[\frac{-1*(-2)*(-3)}{(x-1)^4} +\frac{(-1*2)(-2*2)(-3*2)}{(2x+1)^4} ]$

Similarly, differentiating nth time, we get,

=> $\frac{d^ny}{dx^n}$ =  $\frac{1}{3}[\frac{-1*(-2)*(-3). . .(-n)}{(x-1)^{n+1}} +\frac{(-1*2)(-2*2)(-3*2)...(-n*2)}{(2x+1)^{n+1}} ]$

          =  $\frac{1}{3}[\frac{(-1)^n*n!}{(x-1)^{n+1}} +\frac{(-1)^n*2^n*n!}{(2x+1)^{n+1}} ]$

Therefore, the nth derivative of $\frac{x}{(x-1)(2x+1)}$ is  $\frac{1}{3}[\frac{(-1)^n*n!}{(x-1)^{n+1}} +\frac{(-1)^n*2^n*n!}{(2x+1)^{n+1}} ]$