Question

Find the n th derivative of x^2/(2x+1)(2x+3) (solution in terms of n).

Answer 1

Given:

$\bold{f(x)=\frac{ x^2}{(2x+1)(2x+3)}}$

To find:

Find the nth derivative.

Solution:

Formula:

$\bold{\frac{d}{dx}(\frac{f}{g})=\frac{f'\cdot g-g'\cdot f}{g^2}}$

first derivative:

$\to f'(x)=\frac{d}{dx} \frac{ x^2}{(2x+1)(2x+3)}$ $= \frac{x^2}{(4x^2+8x+3)}$

                           $=\frac{(4x^2+8x+3) 2x-(x^2(8x+8)}{(4x^2+8x+3)^2}\\\\=\frac{(8x^3+16x^2+16x)-(8x^3+8x^2)}{(4x^2+8x+3)^2}\\\\=\frac{(8x^3+16x^2+16x-8x^3-8x^2)}{(4x^2+8x+3)^2}\\\\=\frac{(8x^2+16x)}{(4x^2+8x+3)^2}$

second derivative:

$\to f''(x)= \frac{d}{dx}\frac{(8x^2+16x)}{(4x^2+8x+3)^2} =\frac{(16x+16)(4x^2+8x+3)^2-(2(4x^2+8x+3)(8x+8)(8x^2+16x))}{((4x^2+8x+3)^2)^2}$

 $=\frac{(4x^2+8x+3)((16x+16)(4x^2+8x+3)-(2(4x^2+8x+3)(8x+8)(8x^2+16x)))}{(4x^2+8x+3)^4}\\\\=\frac{((16x+16)(4x^2+8x+3)-(2(4x^2+8x+3)(64x^3+128x^2+64x^2+128x)))}{(4x^2+8x+3)^3}\\\\=\frac{((64x^3+128x^2+48x+64x^2+128x+48)-(2(4x^2+8x+3)(64x^3+128x^2+64x^2+128x)))}{(4x^2+8x+3)^3}\\\\=\frac{((64x^3+192x^2+176x+48)-(2(4x^2+8x+3)(64x^3+192x^2+128x)))}{(4x^2+8x+3)^3}$

$=\frac{((64x^3+192x^2+176x+48)-(2(256x^5+768x^4+512+512x^4+1536x^3+1024x^2+192x^3+576x^2+384x))}{(4x^2+8x+3)^3}$$=\frac{((64x^3+192x^2+176x+48)-(2(256x^5+1280x^4+1728x^3+1600x^2+384x+512))}{(4x^2+8x+3)^3}\\\\=\frac{((64x^3+192x^2+176x+48-(512x^5+2560x^4+3456x^3+3200x^2+786x+1024))}{(4x^2+8x+3)^3}\\\\=\frac{(64x^3+192x^2+176x+48-512x^5-2560x^4-3456x^3-3200x^2-786x-1024)}{(4x^2+8x+3)^3}\\\\=\frac{(-512x^5-2560x^4-3392x^3-3008x^2-610x-976)}{(4x^2+8x+3)^3}\\\\=\frac{-(512x^5+2560x^4+3392x^3+3008x^2+610x+976)}{(4x^2+8x+3)^3}$

Third derivative:

$\to f'''(x)=\frac{-(512x^5+2560x^4+3392x^3+3008x^2+610x+976)}{(4x^2+8x+3)^3}$

$= \frac{-(2560x^4+10240x^3+10176x^2+6016x+610)(4x^2+8x+3)^3}{((4x^2+8x+3)^3)^2}$ $+ \frac{3((4x^2+8x+3)^2(8x+8)(512x^5+2560x^4+3392x^3+3008x^2+610x+976))}{((4x^2+8x+3)^3)^2}$

after solving the value its value:

$=-\frac{2\left(-1024x^6-6144x^5-6272x^4-8704x^3-2868x^2-7568x-10797\right)}{\left(4x^2+8x+3\right)^4}$

Fourth derivative:

$\to f''''(x)=\frac{d}{dx}-\frac{2\left(-1024x^6-6144x^5-6272x^4-8704x^3-2868x^2-7568x-10797\right)}{\left(4x^2+8x+3\right)^4}$

$-2\frac{d}{dx}\left(\frac{-1024x^6-6144x^5-6272x^4-8704x^3-2868x^2-7568x-10797}{\left(4x^2+8x+3\right)^4}\right)$

$= -2\cdot \frac{(-6144x^5-30720x^4-25088x^3-26112x^2-5736x-7568)(4x^2+8x+3)^4}{((4x^2+8x+3)^4)^2}$

$- \frac{4(4x^2+8x+3)^3(8x+8)(-1024x^6-6144x^5-6272x^4-8704x^3-2868x^2-7568x-10797)}{((4x^2+8x+3)^4)^2}$

simplify the value:

$=\frac{16\left(-1024x^7-7168x^6-4096x^5-10240x^4-7900x^3-22432x^2-63741x-40350\right)}{\left(4x^2+8x+3\right)^5}$

n th term is:

$f^n( \frac{x^2}{(4x^2+8x+3)})= \frac{nx^n-1(4x^2+8x+3) -x^2(\frac{d}{dx}(4x^2+8x+3))}{(4x^2+8x+3)^n}$