​ جد المشتقة الثانية اذا علمت ان جذر اكس - جذر واي = 1

Second approach:

$\sqrt{x}-\sqrt{y} = 1\\ \sqrt y = \sqrt x - 1\\ y = x - 2\sqrt x + 1\\ y' = 1-x^{-1/2}\\ y''=\dfrac{1}{2}x^{-3/2}=\dfrac{1}{2x\sqrt x}$

Found out that Melody missed the very last step....... knowing that sqrt x - sqrt y = 1, should have changed the numerator to 1.

1=جد المشتقة الثانية اذا علمت ان جذر اكس - جذر واي 

I asked my Algerian friend, Majid, to translate this for me.  

I think this is correct but I would like another mathematician to check it please.

 Find the second derivative knowing that square root(x) - squre root(y) = 1

$\sqrt{x}-\sqrt{y}=1\\ x^{0.5}-y^{0.5}=1\\ 0.5x^{-0.5}-0.5y^{-0.5}\frac{dy}{dx}=0\\ 0.5x^{-0.5}=0.5y^{-0.5}\frac{dy}{dx}\\ \frac{0.5x^{-0.5}}{0.5y^{-0.5}}=\frac{dy}{dx}\\ \frac{dy}{dx}=\frac{x^{-0.5}}{y^{-0.5}}\\ \frac{dy}{dx}=\frac{y^{0.5}}{x^{0.5}}\\ \frac{d^2y}{dx^2}=\frac{x^{0.5}*0.5y^{-0.5}\frac{dy}{dx}\;-\;y^{0.5}*0.5x^{-0.5}}{x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2y^{0.5}}\frac{dy}{dx}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2y^{0.5}}\frac{y^{0.5}}{x^{0.5}}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ \dfrac{d^2y}{dx^2}=\dfrac{\frac{x^{0.5}}{2x^{0.5}}- \frac{y^{0.5}}{2x^{0.5}} } {x}\\ $

$\dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{0.5}}       \div x\\ \dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{0.5}*x}       \\ \dfrac{d^2y}{dx^2}=\dfrac{x^{0.5}-y^{0.5}}{2x^{1.5}}       \\ \dfrac{d^2y}{dx^2}=\dfrac{\sqrt{x}-\sqrt{y}}{2x\sqrt{x}}       \\ $

Find the derivative of the following via implicit differentiation:
d/dx(sqrt(x) - sqrt(y)) = d/dx(1)
Differentiate the sum term by term and factor out constants:
d/dx(sqrt(x)) - d/dx(sqrt(y)) = d/dx(1)
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/2: d/dx(sqrt(x)) = d/dx(x^(1/2)) = x^(-1/2)/2:
-(d/dx(sqrt(y))) + 1/(2 sqrt(x)) = d/dx(1)
Using the chain rule, d/dx(sqrt(y)) = ( dsqrt(u))/( du) 0, where u = y and ( d)/( du)(sqrt(u)) = 1/(2 sqrt(u)):
1/(2 sqrt(x)) - (d/dx(y))/(2 sqrt(y)) = d/dx(1)
The derivative of y is y'(x):
1/(2 sqrt(x)) - y'(x) 1/(2 sqrt(y)) = d/dx(1)
The derivative of 1 is zero:
1/(2 sqrt(x)) - (y'(x))/(2 sqrt(y)) = 0
Subtract 1/(2 sqrt(x)) from both sides:
-(y'(x))/(2 sqrt(y)) = -1/(2 sqrt(x))
Divide both sides by -1/(2 sqrt(y)):
Answer: |y'(x) = sqrt(y) / sqrt(x)