Question
Differentiate the following functions with respect to x:
$\log(\text{x}+\sqrt{\text{x}^2+1})$
$\log(\text{x}+\sqrt{\text{x}^2+1})$
✓
Answer
Let $\text{y}=\log(\text{x}+\sqrt{\text{x}^2+1})$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dy}}\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\big(\text{x}^2+1\big)^\frac{1}{2}\Big)$
[Using chain rule]
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2}\big(\text{x}^2+1\big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+1\big)\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2\sqrt{\text{x}^2+1}}\times2\text{x}\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big]$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)\Big)=\frac{1}{\sqrt{\text{x}^2+1}}$
Differentiate with respect to x,
$\frac{\text{dy}}{\text{dt}}=\frac{\text{d}}{\text{dy}}\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\frac{\text{d}}{\text{dx}}\Big(\text{x}+\big(\text{x}^2+1\big)^\frac{1}{2}\Big)$
[Using chain rule]
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2}\big(\text{x}^2+1\big)^{\frac{1}{2}-1}\frac{\text{d}}{\text{dx}}\big(\text{x}^2+1\big)\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[1+\frac{1}{2\sqrt{\text{x}^2+1}}\times2\text{x}\Big]$
$=\frac{1}{\text{x}+\sqrt{\text{x}^2+1}}\Big[\frac{\sqrt{\text{x}^2+1}+\text{x}}{\sqrt{\text{x}^2+1}}\Big]$
$=\frac{1}{\sqrt{\text{x}^2+1}}$
So,
$\frac{\text{d}}{\text{dx}}\Big(\log\big(\text{x}+\sqrt{\text{x}^2+1}\big)\Big)=\frac{1}{\sqrt{\text{x}^2+1}}$
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