<s>[INST] How to take the derivative of $Y=\log(x+\sqrt{a^2+x^2})$? 
$$Y=\log(x+\sqrt{a^2+x^2})$$ Find $\dfrac {dY}{dx}$.

My answer:
$$\dfrac 1{x+\sqrt{a^2+x^2}}\cdot\frac{d}{dx}\sqrt{a^2+x^2})$$
Which again goes to a chain rule as $\frac{2}{3} \sqrt{a^2+x^2}\cdot \frac{d}{dx} (a^2+x^2)$
Is there an alternative way other than this? Hints, please.
 [/INST] I think you missed some terms $$Y=\log(x+\sqrt{a^2+x^2})$$ Let $f=x+\sqrt{a^2+x^2}$ which makes $Y=\log(f)$. So $$\frac{dY}{dx}=\frac{dY}{df} \times \frac{df}{dx}=\frac 1 f\times \left(1+\frac{x}{\sqrt{a^2+x^2}}\right)=\frac 1 f\times \left(\frac{x+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}}\right)$$ Replacing $f$ by its expression finally leads to $$\frac{dY}{dx}=\frac{1}{\sqrt{a^2+x^2}}$$
</s>