Question

please solve the sum​

Answer 1

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

Given function is

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

Let we first resolve this in to partial fraction.

Let assume that

$\rm :\longmapsto\: \dfrac{ {x}^{2} }{ {(x + 1)}^{2} (x + 2)} = \dfrac{a}{x + 1} + \dfrac{b}{ {(x + 1)}^{2} } + \dfrac{c}{x + 2} - - - (1)$

On taking LCM, we get

$\rm :\longmapsto\: {x}^{2} = a(x + 1)(x + 2) + b(x + 2) + c {(x + 1)}^{2}$

On substituting x = - 1, we get

$\rm :\longmapsto\: {( - 1)}^{2} = b( - 1 + 2)$

$\bf\implies \:b = 1$

On substituting x = - 2, we get

$\rm :\longmapsto\: {( - 2)}^{2} = c( - 2 + 1) {}^{2}$

$\rm :\longmapsto\:4 = c {( - 1)}^{2}$

$\bf\implies \:c = 4$

On substituting x = 0, we get

$\rm :\longmapsto\: {0}^{2} = a(0 + 1)(0 + 2) + b(0 + 2) + c {(0 + 1)}^{2}$

$\rm :\longmapsto\:0 = 2a + 2b + c$

On substituting the values of b and c, we get

$\rm :\longmapsto\:0 = 2a + 2 + 4$

$\rm :\longmapsto\:0 = 2a + 6$

$\rm :\longmapsto\:2a = - 6$

$\bf\implies \:a = - \: 3$

On substituting the values of a, b and c in equation (1), we get

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

So,

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

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}y = - 3\dfrac{d}{dx} \dfrac{1}{x + 1} +\dfrac{d}{dx} \dfrac{1}{ {(x + 1)}^{2} } +4\dfrac{d}{dx} \dfrac{1}{x + 2}$

We know,

$\red{\boxed{ \bf{ \dfrac{d}{dx} \frac{1}{ {x}^{n} } = \frac{ - n}{ {x}^{n + 1} }}}}$

So, using this result, we get

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{( - 3)( - 1)}{ {(x + 1)}^{2} } + \dfrac{( - 2)}{ {(x + 1)}^{3} } + \dfrac{4( - 1)}{ {(x + 2)}^{2} }$

can be rewritten as

$\rm :\longmapsto\: \dfrac{dy}{dx} = \dfrac{3}{ {(x + 1)}^{2} } - \dfrac{2}{ {(x + 1)}^{3} } - \dfrac{4}{ {(x + 2)}^{2} }$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}\dfrac{dy}{dx} =\dfrac{d}{dx} \dfrac{3}{ {(x + 1)}^{2} } - \dfrac{d}{dx} \dfrac{2}{ {(x + 1)}^{3} } -\dfrac{d}{dx} \dfrac{4}{ {(x + 2)}^{2} }$

$\rm :\longmapsto\: \dfrac{d^{2} y}{dx ^{2} } = \dfrac{3( - 2)}{ {(x + 1)}^{3} } - \dfrac{2( - 3)}{ {(x + 1)}^{4} } - \dfrac{4( - 2)}{ {(x + 2)}^{3} }$

$\rm :\longmapsto\: \dfrac{d^{2} y}{dx ^{2} } = \dfrac{ - 6}{ {(x + 1)}^{3} } + \dfrac{6}{ {(x + 1)}^{4} } + \dfrac{8}{ {(x + 2)}^{3} }$

Additional Information :-

$\boxed{ \rm{ \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}}$

$\boxed{ \rm{ \dfrac{d}{dx}k = 0}}$

$\boxed{ \rm{ \dfrac{d}{dx}x = 1}}$

$\boxed{ \rm{ \dfrac{d}{dx} \sqrt{x} = \frac{1}{2 \sqrt{x} }}}$

$\boxed{ \rm{ \dfrac{d}{dx}logx = \frac{1}{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} {e}^{x} = {e}^{x}}}$

$\boxed{ \rm{ \dfrac{d}{dx} {a}^{x} = {a}^{x}loga}}$