Question

y=x²/(x-1) ² (x+2).
calculate the nth derivative of y​

Answer 1

Given:

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

To find:

Derivative of y​

Solution:

Formula:

$\frac{d}{dx} \frac{u}{v} = \frac{v \frac{du}{dx}- u \frac{dv}{dx}}{v^2}$

Solving the given value:

$\to y= \frac{x^2}{(x-1)^2(x+2)}\\\\\to y= \frac{x^2}{(x^2+1-2x)(x+2)}\\\\\to y= \frac{x^2}{(x^3+x-2x^2 +2x^2+2-4x)}\\\\\to y= \frac{x^2}{(x^3-3x+2)}$

calculating the Derivative:

$y'= \frac{(x^3-3x+2) \cdot 2x - x^2 \cdot \frac{d}{dx} (x^3-3x+2) }{(x^3-3x+2)^2}\\\\ y'= \frac{(x^3-3x+2) \cdot 2x - x^2 \cdot (3x^2-3) }{(x^3-3x+2)^2}\\\\y'= \frac{(2x^4-6x^2+4x -3x^4+3x^2) }{(x^3-3x+2)^2}\\\\y'= \frac{(-x^4-3x^2+4x) }{(x^3-3x+2)^2}$

calculating second Derivative:$y''= \frac{(x^3-3x+2)^2 (-4x^3-6x+4) - (-x^4 -3x^2+4x) \frac{d}{dx}(x^3-3x+2)^2 }{((x^3-3x+2)^2)^2}\\\\y''= \frac{(x^3-3x+2)^2 \cdot (-4x^3-6x+4) - (-x^4 -3x^2+4x) \cdot 2(x^3-3x+2)( (3x^2-3) }{((x^3-3x+2)^2)^2}\\\\y''= \frac{(x^3-3x+2)^2 \cdot (-4x^3-6x+4) - (-x^4 -3x^2+4x) \cdot 2(x^3-3x+2)( (3x^2-3) }{((x^3-3x+2)^2)^2}$

and so on....

Answer 2

Step-by-step explanation:

It is done by using the by parts integration method.