Question

find the nth derivative of 2x+1/x²–4​

Answer 1

$\large\underline{\sf{Solution-}}$

$\bf :\longmapsto\:Let \: y = \dfrac{2x + 1}{ {x}^{2} - 4}$

$\rm :\longmapsto\:y = \dfrac{2x + 1}{(x + 2)(x - 2)}$

Let we first resolve in to partial fraction.

$\bf :\longmapsto\:Let \: \dfrac{2x + 1}{(x+ 2)(x - 2) } = \dfrac{a}{x + 2}+ \dfrac{b}{x - 2} - - (1)$

$\rm :\longmapsto\:2x + 1 = a(x - 2) + b(x + 2)$

On substituting x = 2, we get

$\rm :\longmapsto\:2 \times 2 + 1 = a(2 - 2) + b(2+ 2)$

$\rm :\longmapsto\:5 = 4b$

$\rm :\longmapsto\:b = \dfrac{5}{4} - - - (2)$

On substituting x = - 2, we get

$\rm :\longmapsto\:2( - 2) + 1 = a( - 2 - 2) + b( - 2 + 2)$

$\rm :\longmapsto\: - 4 + 1 = a( -4)$

$\rm :\longmapsto\: - 3 = a( -4)$

$\rm :\longmapsto\:a = \dfrac{3}{4} - - - (3)$

On substituting equation (2) and (3), in equation (1),

$\bf :\longmapsto\:\: \dfrac{2x + 1}{(x+ 2)(x - 2) } = \dfrac{ \frac{3}{4} }{x + 2}+ \dfrac{ \frac{5}{4} }{x - 2}$

So it implies,

$\bf :\longmapsto\:\: \dfrac{2x + 1}{ {x}^{2} - 4} = \dfrac{3}{4(x + 2)}+ \dfrac{5}{4(x - 2)}$

Hence,

Differentiating n times, we get

$\bf :\longmapsto\:\: \dfrac{ {d}^{n} }{d {x}^{n} }\dfrac{2x + 1}{ {x}^{2} - 4} = \dfrac{ {d}^{n} }{d {x}^{n} } \dfrac{3}{4(x + 2)}+ \dfrac{ {d}^{n} }{d {x}^{n} }\dfrac{5}{4(x - 2)}$

$\rm \: \: = \: \dfrac{3}{4}\dfrac{ {( - 1)}^{n}n !}{ {(x + 2)}^{n + 1} } + \dfrac{5}{4}\dfrac{ {( - 1)}^{n}n !}{ {(x - 2)}^{n + 1} }$

$\green{\boxed{ \bf{ \: \: \: \because \: \:\dfrac{ {d}^{n} }{d {x}^{n} }\dfrac{1}{ax + b} = \dfrac{ {( - 1)}^{n} {a}^{n} n !}{ {(ax + b)}^{n + 1} } \: \: \: \: \: \: \: \: }}}$