Question

Show that the function is always increasing

$f(x) = log(1 + x) - \frac{2x}{x + 2}$
​

Answer 1

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

Given function is

$\rm \: f(x) = log(1 + x) - \dfrac{2x}{x + 2}$

Let first define the domain of f(x).

Now, log(1 + x) is defined when x + 1 > 0

Now, Consider

$\rm \: f(x) = log(1 + x) - \dfrac{2x}{x + 2}$

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

$\rm \: \dfrac{d}{dx}f(x) =\dfrac{d}{dx}\bigg[ log(1 + x) - \dfrac{2x}{x + 2}\bigg]$

We know,

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

and

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} = \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}$

So, using these results, we get

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

We know,

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

and

$\boxed{\tt{ \dfrac{d}{dx}k = 0 \: }}$

So, using this result, we get

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

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

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

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

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

$\rm \: f'(x) = \dfrac{ {x}^{2} }{(x + 1) {(x + 2)}^{2} }$

Now, as

$\rm \: {x}^{2} \geqslant 0$

$\rm \: {(x + 2)}^{2} > 0$

$\rm \: x + 1 > 0$

So,

$\rm\implies \:\rm \: f'(x) = \dfrac{ {x}^{2} }{(x + 1) {(x + 2)}^{2} } \geqslant 0$

$\rm\implies \:\rm \: f'(x) \: is \: always \: increasing \: when \: x > - 1$

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ADDITIONAL INFORMATION

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