TITLE: show that two equation are the same in differentiation
QUESTION [0 upvotes]: If $y=(x+\sqrt{x^2+1})^2$, show that $\frac{dy}{dx} =$ $\dfrac{2y}{\sqrt{x^2+1}}$ 
Then we have let $ u=(x+\sqrt{x^2+1})$
$$
\frac{dy}{du} = 2(x\sqrt{x^2+1})$$ and $\qquad \frac{du}{dx} =\sqrt{x^2} \quad then\quad x.
$
however I can't show that $\frac{dy}{dx} =$ $\dfrac{2y}{\sqrt{x^2+1}}$
please help me out.
Thanks in advance.

REPLY [2 votes]: $$y'=2(x+\sqrt{x^2+1})(x+\sqrt{x^2+1})'$$
$$y'=2(x+\sqrt{x^2+1})\left(1+\frac{1}{2\cdot \sqrt{x^2+1}}\cdot (x^2+1)'\right)$$
$$y'=2(x+\sqrt{x^2+1})\left(\frac{x+\sqrt{x^2+1}}{ \sqrt{x^2+1}}\right)$$
$$y'=2\cdot \frac{\left(x+\sqrt{x^2+1}\right)^2}{ \sqrt{x^2+1}}$$
$$y'=\frac{2\cdot y}{\sqrt{x^2+1}}$$
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