Question

find dy/dx if log(x^2+2x)​

Answer 1

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

Given function is

$\rm :\longmapsto\ \: \: log( {x}^{2} + 2x)$

Let us assume that

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

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

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

We know,

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

and

$\boxed{\tt{ \dfrac{d}{dx}f[g(x)] = f'[g(x)]\dfrac{d}{dx}g(x)}}$

So, using this, we get

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

We know,

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

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ {x}^{2} + 2x}\bigg[2x + 2\bigg]$

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

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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

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

Given function is

$\rm :\longmapsto\ \: \: log( {x}^{2} + 2x)$

Let us assume that

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

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

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

We know,

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

and

$\boxed{\tt{ \dfrac{d}{dx}f[g(x)] = f'[g(x)]\dfrac{d}{dx}g(x)}}$

So, using this, we get

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

We know,

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

So, using this, we get

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{ {x}^{2} + 2x}\bigg[2x + 2\bigg]$

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information

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