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?

$$\frac{d}{{dx}}(\frac{u}{v}) = \frac{{u'v - v'u}}{{v^2 }}$$

 

[iMAGICIALoTV]

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