Second Derivative of Square Root Equation

[Sakha]

Homework Statement

\sqrt{x}+\sqrt{y}= 4 prove that y''=\frac{2}{x\sqrt{x}}

The Attempt at a Solution

\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}

I tried solving for the second derivative and got

\frac{d^2y}{dx^2}=-\frac{-x\frac{dy}{dx}-y}{2\sqrt{\frac{y}{x}}x^2}
[
Which is wrong if I plug in values.
Anyone sees my error?

 

[danago]

Homework Statement

\sqrt{x}+\sqrt{y}= 4 prove that y''=\frac{2}{x\sqrt{x}}

The Attempt at a Solution

\frac{dy}{dx}=-\frac{\sqrt{y}}{\sqrt{x}}

I tried solving for the second derivative and got

\frac{d^2y}{dx^2}=-\frac{-x\frac{dy}{dx}-y}{2\sqrt{\frac{y}{x}}x^2}
[
Which is wrong if I plug in values.
Anyone sees my error?

Your first derivative looks ok to me, but I am not too sure where your second derivative came from. Did you use the quotient rule?

<br /> \frac{d}{{dx}}(\frac{u}{v}) = \frac{{u&#039;v - v&#039;u}}{{v^2 }}<br />

 

[iMAGICIALoTV]

How did you get \frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}?