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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY5 Marks
Question
Differentiate the following functions with respect to x:
$\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
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Answer

Let $\text{y}=\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}$
Differentiate it with respect to x,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]$
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]}\frac{\text{d}}{\text{dx}}\Big[\text{x}+2+\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\Big]$
[Using chain rule]
$=\frac{1}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}\times\Big[1+0+\frac{1}{2}\big(\text{x}^2+4\text{x}+1\big)^\frac{1}{2}\frac{\text{d}}{\text{dx}}(\text{x}^2+4\text{x}+1)\Big]$
$=\frac{1+\frac{(2\text{x}+4)}{2\big(\sqrt{\text{x}^2+4\text{x}+1}\big)}}{\big[\text{x}+2+\sqrt{\text{x}^4+4\text{x}+1}\big]}$
$=\frac{\sqrt{\text{x}^2+4\text{x}+1}+\text{x}+2}{\big[\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big]\times\sqrt{\text{x}^2+4\text{x}+1}}$
$=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big[\log\big\{\text{x}+2+\sqrt{\text{x}^2+4\text{x}+1}\big\}\Big]=\frac{1}{\sqrt{\text{x}^2+4\text{x}+1}}$

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