Question

Find the interval of increasing and decreasing

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

Answer 1

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

Given function is

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

Let first define the domain of f(x).

$\boxed{ \tt{ \: x + 1 > 0 \: \rm \implies\:x > - 1 \: }}$

Now, Consider

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

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

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

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, we get

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

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

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

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

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

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

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

As x + 1 > 0

$\bf\implies \:f'(x) \geqslant 0$

$\bf\implies \:f(x) \: is \: increasing \: \forall \: x \in \: ( - 1, \: \infty )$

More to know :

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