Chain rule with$~\left(\frac{1}{2}\ln\left(\frac{x^{2}+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^{2}} \right)\right)$?

Question

Please jump to the bottom of this post if you just want to see my current critical problem.

$$y=\frac{1}{2\sqrt{2}}\left(\frac{1}{2}\ln\left(\frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1}\right)+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^{2}} \right)\right)\tag{1}$$

I want to derive equation of$~\frac{dy}{dx}~$

I know this problem can be solved using chain rule.

My attempt is as following

$$f\left(x\right):=\frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1}\tag{2}$$

$$=\frac{\left(x^{2}+\sqrt{2}x+1\right)}{\left(x^{2}-\sqrt{2}x+1\right)}\frac{\left(x^{2}+\sqrt{2}x+1\right)}{\left(x^{2}+\sqrt{2}x+1\right)}$$

$$=\frac{\left(x^{2}+\sqrt{2}x+1\right)^{2}}{\left(x^{2}-\sqrt{2}x+1\right)\left(x^{2}+\sqrt{2}x+1\right)}$$

$$=\frac{\left(x^{2}+1+\sqrt{2}x\right)^{2}}{\left(\left(x^{2}+1\right)-\sqrt{2}x\right)\left(\left(x^{2}+1\right)+\sqrt{2}x\right)}$$

$$=\frac{\left(\left(x^{2}+1\right)+\sqrt{2}x\right)^{2}}{\left(\left(x^{2}+1\right)-\sqrt{2}x\right)\left(\left(x^{2}+1\right)+\sqrt{2}x\right)}$$

$$=\frac{\left(\left(x^{2}+1\right)+\sqrt{2}x\right)^{2}}{\left(\left(x^{2}+1\right)^{2}-2x^{2}\right)}$$

$$=\frac{\left(\left(x^{2}+1\right)+\sqrt{2}x\right)^{2}}{\left(x^{4}+2x^{2}+1-2x^{2}\right)}$$

$$=\frac{\left(\left(x^{2}+1\right)+\sqrt{2}x\right)^{2}}{\left(x^{4}+1\right)}$$

$$=\frac{\left(\left(x^{2}+1\right)^{2}+2\sqrt{2}\left(x^{2}+1\right)x+2x^{2}\right)^{}}{\left(x^{4}+1\right)}$$

$$=\frac{\left(\left(x^{4}+2x^{2}+1\right)^{}+2\sqrt{2}\left(x^{3}+x\right)+2x^{2}\right)^{}}{\left(x^{4}+1\right)}$$

$$=\frac{\left(x^{4}+2\sqrt{2}x^{3}+4x^{2}+2\sqrt{2}x+1\right)}{\left(x^{4}+1\right)}~~\leftarrow~~\text{Nominator seems really complicated}~~$$

$$y=\frac{1}{2\sqrt{2}}\left(\frac{1}{2}\ln\left(\frac{x^{2}+\sqrt{2}x+1}{x^{2}-\sqrt{2}x+1}\right)+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^{2}} \right)\right)\tag{1}$$

$$g\left(x\right):=\frac{\sqrt{2}x}{1-x^{2}}\tag{2}$$

$$y=\frac{1}{2\sqrt{2}}\left(\frac{1}{2}\ln\left(f\left(x\right)\right)+\tan^{-1}\left(g\left(x\right)\right)\right)$$

$$\frac{dy}{dx}=\frac{1}{2\sqrt{2}}\left\{\frac{1}{2}\frac{d}{dx}\left(\ln\left(f\left(x\right)\right)\right)+\frac{d}{dx}\left(\tan^{-1}\left(g\left(x\right)\right)\right)\right\}$$

$$=\frac{1}{2\sqrt{2}}\left(\frac{1}{2}\frac{1}{f\left(x\right)}\frac{d}{dx}\left(f\left(x\right)\right)+\frac{d}{dx}\left(\tan^{-1}\left(g\left(x\right)\right)\right)\right)$$

$$=\frac{1}{2\sqrt{2}}\left(\frac{f'\left(x\right)}{2f\left(x\right)}+\frac{g'\left(x\right) }{1+g\left(x\right)^{2} }\right)$$

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

$$=\left(x^2+\sqrt{2}x+1\right)\left(x^2-\sqrt{2}x+1\right)^{-1}$$

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

Answer 1

Hint

Work with pieces and consider $$y=\frac 12 \log(f(x))+\tan^{-1}(g(x))$$ $$y'=\frac 12 \frac{f'(x)}{ f(x)}+\frac{g'(x)}{1+g(x)^2}$$ Now $$f(x)=\frac{x^{2}+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\qquad \text{and} \qquad g(x)=\frac{\sqrt{2}x}{1-x^{2}}$$

Answer 2

If you remember that, $\large\int\frac1{x^4+1}dx=\frac1{2\sqrt2}\left(\frac{1}{2}\ln\left(\frac{x^{2}+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^{2}} \right)\right)+C$
You can simply conclude that $y'=\dfrac1{x^4+1}$.