<s>[INST] Derivative of $\frac{d}{dx}\ln (x+\sqrt{x^2+1})$ Show that $\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1}{\sqrt{x^2+1}}$
So I've done this so far:
$$\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}}$$
I have tried a few combinations but cannot attain the desired result. Could someone offer just a hint to get me going?
Thanks
 [/INST] Using Chain Rule of derivative , $$\frac{d(\ln(x+\sqrt{x^2+1}))}{dx}=\frac{d(\ln(x+\sqrt{x^2+1}))}{d(x+\sqrt{x^2+1})}\frac{d(x+\sqrt{x^2+1})}{dx}$$
$$=\frac1{x+\sqrt{x^2+1}}\left(1+\frac{2x}{2\sqrt{x^2+1}}\right)=\frac{x+\sqrt{x^2+1}}{(x+\sqrt{x^2+1})\sqrt{x^2+1}}=\frac1{\sqrt{x^2+1}}$$
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