Prove the following

Question:

$\log \left(x+\sqrt{x^{2}+a}\right)$

Solution:

Let $y=\log \left(x+\sqrt{x^{2}+a}\right)$

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=\frac{d}{d x} \log \left(x+\sqrt{x^{2}+a}\right)$

$=\frac{1}{x+\sqrt{x^{2}+a}} \cdot \frac{d}{d x}\left(x+\sqrt{x^{2}+a}\right)$

$=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left[1+\frac{1}{2 \sqrt{x^{2}+a}} \times \frac{d}{d x}\left(x^{2}+a\right)\right]$

$=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left[1+\frac{1}{2 \sqrt{x^{2}+a}} \cdot 2 x\right]$

$=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left[1+\frac{x}{\sqrt{x^{2}+a}}\right]$

$=\frac{1}{x+\sqrt{x^{2}+a}} \cdot\left(\frac{\sqrt{x^{2}+a}+x}{\sqrt{x^{2}+a}}\right)=\frac{1}{\sqrt{x^{2}+a}}$

Thus,

$\frac{d y}{d x}=\frac{1}{\sqrt{x^{2}+a}} .$

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