Differentiate --> I'm trying to differentiate but am having a hard time -- any pointers appreciated

Question

here is my poor attempt https://imgur.com/a/fX75 $f(x) = \log \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^1 + 1} + x}$

Answer 1

$$f(x) = \log \frac{\sqrt{x^2 + 1} - x}{\sqrt{x^2 + 1} + x}=$$

$$\log (\sqrt{x^2 + 1} - x)-log(\sqrt{x^2 + 1} + x)$$

$$ \frac {d}{dx} \log (\sqrt{x^2 + 1} - x)= $$

$$\frac {x- \sqrt {x^2+1}}{x^2+1-x \sqrt {x^2+1}} $$

Similarly $$ \frac {d}{dx} \log (\sqrt{x^2 + 1} + x)= $$

$$\frac {x+ \sqrt {x^2+1}}{x^2+1+x \sqrt {x^2+1}} $$

$$ f'(x)=\frac {x- \sqrt {x^2+1}}{x^2+1-x \sqrt {x^2+1}}- \frac {x+ \sqrt {x^2+1}}{x^2+1+x \sqrt {x^2+1}} =2x-2\sqrt {x^2+1}.$$

Answer 2

by the chain and power rule we get $$f'(x)=\frac{\left(x+\sqrt{x+1}\right) \left(\frac{\frac{x}{\sqrt{x^2+1}}-1}{x+\sqrt{x+1}}- \frac{\left(\frac{1}{2 \sqrt{x+1}}+1\right) \left(\sqrt{x^2+1}-x\right)}{\left(x+\sqrt{x+1}\right)^2}\right)}{\sqrt{x^2+1}-x}$$ this Can be simplified

Answer 3

Let $y=\sinh^{-1}x$. Then $x=\sinh y$, so $$1=\cosh yy^{\prime}=\sqrt{1+\sinh^2y}y^{\prime}=\sqrt{1+x^2}y^{\prime}$$ $$y^{\prime}=\frac d{dx}\sinh^{-1}x=\frac1{\sqrt{x^2+1}}$$ On the other hand, $$x=\frac{e^y-e^{-y}}2$$ So $e^{2y}-2xe^y-1=0$. Solving for $e^y$, $$e^y=x\pm\sqrt{x^2+1}=x+\sqrt{x^2+1}$$ Because $e^y>0$. Then $$y=\ln\left(\sqrt{x^2+1}+x\right)=\sinh^{-1}x$$ Also $$\left(\sqrt{x^2+1}+x\right)\left(\sqrt{x^2+1}-x\right)=x^2+1-x^2=1$$ So $$\begin{align}\ln\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x}&=\ln\left(\sqrt{x^2+1}-x\right)-\ln\left(\sqrt{x^2+1}+x\right)\\ &=\ln\left[\left(\sqrt{x^2+1}+x\right)^{-1}\right]-\ln\left(\sqrt{x^2+1}+x\right)\\ &=-2\ln\left(\sqrt{x^2+1}+x\right)=-2\sinh^{-1}x\end{align}$$ So $$\frac d{dx}\ln\frac{\sqrt{x^2+1}-x}{\sqrt{x^2+1}+x}=-2\frac d{dx}\sinh^{-1}x=\frac{-2}{\sqrt{x^2+1}}$$ EDIT It has become clear that what was desired was $$\begin{align}\frac d{dx}\ln\frac{\sqrt{x^2+1}-x}{\sqrt{x+1}+x}&=\frac d{dx}\ln\left(\sqrt{x^2+1}-x\right)-\frac d{dx}\ln\left(\sqrt{x+1}+x\right)\\ &=\frac{\frac{2x}{2\sqrt{x^2+1}}-1}{\sqrt{x^2+1}-x}-\frac{\frac1{2\sqrt{x+1}}+1}{\sqrt{x+1}+x}\\ &=\frac{x-\sqrt{x^2+1}}{\sqrt{x^2+1}\left(\sqrt{x^2+1}-x\right)}-\frac{1+2\sqrt{x+1}}{2\sqrt{x+1}\left(\sqrt{x+1}+x\right)}\\ &=\frac{-1}{\sqrt{x^2+1}}-\frac{\left(1+2\sqrt{x+1}\right)\left(\sqrt{x+1}-x\right)}{2\sqrt{x+1}\left(x+1-x^2\right)}\\ &=\frac{-1}{\sqrt{x^2+1}}-\frac{x+2+(1-2x)\sqrt{x+1}}{2\sqrt{x+1}\left(x+1-x^2\right)}\end{align}$$