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
$\endgroup$ 13 Answers
$\begingroup$Your differentiation is correct, and you applied the chain rule correctly. But as you note, your result can be simplified.
Let's rewrite the fraction, expressing the term $x(x^2 + 1)^{-0.5}$ in the numerator as $\frac{x}{\sqrt {x^2 + 1}}$, then simply multiply the numerator and denominator by $\sqrt{x^2 + 1}$:
$$\begin{align}\frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}} & = \frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\\ \\ & = \frac{\sqrt{x^2 +1} + x}{{(x+\sqrt{x^2+1})\sqrt{x^2+1}}}\\ \\ & =\frac{\color{blue}{\bf x+\sqrt{x^2+1}}}{(\color{blue}{\bf x+\sqrt{x^2+1}})\sqrt{x^2+1}} \\ \\ & =\frac1{\sqrt{x^2+1}}\end{align}$$
$\endgroup$ 3 $\begingroup$Recognize the function $\ln(x+\sqrt{x^2+1})$ as the inverse hyperbolic sine, $\text{arcsinh } x$. According to the inverse function theorem we have
$$\frac{d}{dx} \text{arcsinh } x=\frac{1}{\cosh \text{arcsinh }x}=\frac{1}{\dfrac{e^{\ln(x+\sqrt{x^2+1})}+e^{-(\ln{x+\sqrt{x^2+1}})}}{2}}=\frac{1}{\sqrt{x^2+1}}. $$
$\endgroup$ $\begingroup$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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