Question
Differentiate the following w.r.t. x:
$\log\Big(\text{x}+\sqrt{\text{x}^2+\text{a}}\Big)$
$\log\Big(\text{x}+\sqrt{\text{x}^2+\text{a}}\Big)$
✓
Answer
Let $\text{y}=\log\Big(\text{x}+\sqrt{\text{x}^2+\text{a}}\Big)$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\frac{\text{d}}{\text{dx}}\Big[\text{x}+\sqrt{\text{x}^2+\text{a}}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\Big[1+\frac{1}{2}(\text{x}^2+\text{a})^{\frac{-1}{2}}\cdot2\text{x}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}}}\Big)$
$-\frac{\big(\sqrt{\text{x}^2+\text{a}}+\text{x}\big)}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)\big(\sqrt{\text{x}^2+\text{a}}\big)}$
$=\frac{1}{\big(\sqrt{\text{x}^2+\text{a}}\big)}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\frac{\text{d}}{\text{dx}}\Big[\text{x}+\sqrt{\text{x}^2+\text{a}}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\Big[1+\frac{1}{2}(\text{x}^2+\text{a})^{\frac{-1}{2}}\cdot2\text{x}\Big]$
$=\frac{1}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)}\cdot\Big(1+\frac{\text{x}}{\sqrt{\text{x}^2+\text{a}}}\Big)$
$-\frac{\big(\sqrt{\text{x}^2+\text{a}}+\text{x}\big)}{\big(\text{x}+\sqrt{\text{x}^2+\text{a}}\big)\big(\sqrt{\text{x}^2+\text{a}}\big)}$
$=\frac{1}{\big(\sqrt{\text{x}^2+\text{a}}\big)}$
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