Question

find the derivatives of log(x^2+x+2/x^2-x+2)​

Answer 1

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

Given function is

$\rm \: log\bigg(\dfrac{ {x}^{2} + x + 2}{ {x}^{2} - x + 2} \bigg)$

Let assume that

$\rm \: y \: = \: log\bigg(\dfrac{ {x}^{2} + x + 2}{ {x}^{2} - x + 2} \bigg)$

We know,

$\boxed{\tt{ \: log \frac{x}{y} = logx - logy \: }}$

So, using this result, we have

$\rm \: y = log( {x}^{2} + x + 2) - log( {x}^{2} - x + 2)$

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

$\rm \:\dfrac{d}{dx} y =\dfrac{d}{dx} log( {x}^{2} + x + 2) -\dfrac{d}{dx} log( {x}^{2} - x + 2)$

We know,

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

So, using this result, we have

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

We know,

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

So, using this result, we get

$\rm \: \dfrac{dy}{dx} = \dfrac{1}{ {x}^{2} + x + 2}( 2x + 1 + 0) - \dfrac{1}{ {x}^{2} - x + 2}( 2x - 1 + 0)$

$\rm \: \dfrac{dy}{dx} = \dfrac{2x + 1}{ {x}^{2} + x + 2} - \dfrac{2x - 1}{ {x}^{2} - x + 2}$

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

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

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

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

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

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

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ADDITIONAL INFORMATION

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$

Answer 2

REFER THE GIVEN ATTACHMENTS