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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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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? –  Bidit Acharya 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

1 Answer 1

up vote 3 down vote accepted

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

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

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

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

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

share|improve this answer
    
very close but should be $\dfrac{2y}{\sqrt{x^2+1}}$ –  Sb Sangpi Apr 4 '12 at 12:39
1  
@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

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