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.