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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.

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    http://lmgtfy.com/?q=chain+rule2012-04-04

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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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    @SbSangpi [Chain rule](http://en.wikipedia.org/wiki/Chain_rule)2012-04-04