#### Provide solution for RD Sharma maths class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 26

f(x) is increasing in (0,$\infty$)

and f(x) is decreasing (-1,0)

Given:

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

To find:

We have to find the intervals in which f(x) is increasing and decreasing.

Hint:

First, we find critical point then use property of increasing and decreasing.

Solution:

We have,

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

Differentiating w.r.t. x we get,

\begin{aligned} &f^{\prime}(x)=\frac{d}{d x}\left\{\log (1+x)-\frac{x}{1+x}\right\} \\ &f^{\prime}(x)=\frac{1}{1+x}-\left[\frac{(1+x)-x}{(1+x)^{2}}\right] \end{aligned}

\begin{aligned} &=\frac{1}{1+x}-\frac{1}{(1+x)^{2}},\left[\therefore \frac{d}{d x}\left(\frac{u}{v}\right)=\frac{v \frac{d u}{d x} \cdot u \frac{d v}{d x}}{v^{2}}\right] \\ &=\frac{x}{(1+x)^{2}} \end{aligned}

For critical points. We must have,

\begin{aligned} &f^{\prime}(x)=0\\ &\Rightarrow \frac{x}{(1+x)^{2}}=0\\ &\Rightarrow x=0 \text { and domain of }(1+x)^{2} \text { is }(-1, \infty)\\ &\text { Clearly, } f^{\prime}(x)>0 \text { if } x>0\\ &\text { and } f^{\prime}(x)<0 \text { if }-1

Hence,f(x) is increasing in (0,$\infty$), decreases in (-1,0).