# show that two equation are the same in differentiation

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 $\frac{dy}{dx} =$ $\dfrac{2y}{\sqrt{x^2+1}}$

-
what's your u ? – Glougloubarbaki Apr 4 '12 at 12:25
Yes, I am assuming that you missed out $u$. (Is it that $u=x\sqrt{x^2+1} ~?$) Either ways, why don't you use chain rule, to differentiate $y$ straight-away? – funktor Apr 4 '12 at 12:35
can u plz give me the chain rule formula? thx – Sb Sangpi Apr 4 '12 at 13:01
lmgtfy.com/?q=chain+rule – Glougloubarbaki Apr 4 '12 at 14:30

$$y'=2(x+\sqrt{x^2+1})(x+\sqrt{x^2+1})'$$

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

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

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

$$y'=\frac{2\cdot y}{\sqrt{x^2+1}}$$

-
very close but should be $\dfrac{2y}{\sqrt{x^2+1}}$ – Sb Sangpi Apr 4 '12 at 12:39
@SbSangpi isn't it ? – pedja Apr 4 '12 at 12:40
Can u plz write the chain rule formula, so that I can use next time. thx – Sb Sangpi Apr 4 '12 at 13:14
@SbSangpi Chain rule – pedja Apr 4 '12 at 13:18