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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Show that $\text{y} = \log (1+\text{x)}-\frac{2\text{x}}{2+\text{x}},\text{x}>-1$ is an increasing function of x throughout its domain.
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Answer

Given: $\text{y} = \log(1+\text{x)}-\frac{2\text{x}}{2+\text{x}}$
$\therefore \ \frac{\text{dy}}{\text{dx}}=\frac{1}{1+\text{x}}\frac{\text{d}}{\text{dx}}(1+\text{x})$ $-\Bigg[\frac{(2 + \text{x)}\frac{\text{d}}{\text{dx}} (2\text{x}) -2\text{x} \frac{\text{d}}{\text{dx}}(2 + \text{x)}}{(2 + \text{x)}^{2}}\Bigg]$
$=\frac{1}{1+\text{x}}-\bigg[\frac{(2+\text{x)}2-2\text{x}}{(2+\text{x)}^2}\bigg]=\frac{1}{1+\text{x}}$ $-\frac{(4+2\text{x}-2\text{x})}{(2+\text{x)}^2}=\frac{1}{1+\text{x}}-\frac{4}{(2+\text{x)}^2}$
$\Rightarrow \ \frac{\text{dy}}{\text{dx}}=\frac{(2+\text{x)}^2-4(1+\text{x)}}{(1+\text{x)}(2+\text{x)}^2}$ $=\frac{\text{x}^2}{(1+\text{x)}(2+\text{x)}^2}\ \dots\dots\text{(i)}$
Domain of the given function is given to be $\text{x}>-1 \Rightarrow\ \text{x}+1>0$
$\text{Alos}\ (2+\text{x)}^2>0\ \text{and }\text{x}^2\geq0$
$\therefore\ \text{From eq.(i)},\frac{\text{dy}}{\text{dx}}\geq\ 0$ for all x in domain x > -1 and f is an increasing function.

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