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If $y=\log \left(x+\sqrt{x^2+a}\right)$, then find $\frac{d y}{d x}$.
Solution:
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Verified Answer
$\therefore \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}+\sqrt{\mathrm{x}^2+\mathrm{a}}\right)$ $=\frac{1}{\left(x+\sqrt{x^2+a}\right)} \cdot \frac{d}{d x}\left[x+\sqrt{x^2+a}\right]$
$$
=\frac{1}{\left(x+\sqrt{\left.x^2+1\right)}\right.}\left[1+\frac{1}{2}\left(x^2+a\right)^{-1 / 2} \cdot 2 x\right]
$$
$$
=\frac{1}{x+\sqrt{x^2+a}} \cdot \int 1+\frac{x}{\sqrt{x^2+a}}
$$
$$
=\frac{\left(\sqrt{x^2+a}+x\right)}{\left(x+\sqrt{x^2+a}\right)\left(\sqrt{x^2+a}\right)}=\frac{1}{\left(\sqrt{x^2+a}\right)}
$$
$$
=\frac{1}{\left(x+\sqrt{\left.x^2+1\right)}\right.}\left[1+\frac{1}{2}\left(x^2+a\right)^{-1 / 2} \cdot 2 x\right]
$$
$$
=\frac{1}{x+\sqrt{x^2+a}} \cdot \int 1+\frac{x}{\sqrt{x^2+a}}
$$
$$
=\frac{\left(\sqrt{x^2+a}+x\right)}{\left(x+\sqrt{x^2+a}\right)\left(\sqrt{x^2+a}\right)}=\frac{1}{\left(\sqrt{x^2+a}\right)}
$$
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