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ok this is implicit differentiation.
you want to take the derivative with respect to x.
d/(dx) of x^(1/2) is 1/(2(x^(1/2)) one over the quantity two square root of x.
d/(dx) of y^(1/2) is 1/(2(x^(1/2)) dy/dx
d/(dx) of 2 is a constant so it goes to 0.
So your first equation for dy/dx = 1/(2(x^(1/2)) + 1/(2(y^(1/2)) dy/dx = 0
I'll do the second derivative d^2y/dx^2 in the next post
1/(2(y^(1/2)) dy/dx = -(1/(2(x^(1/2)) )
dy/dx = -(1/(2(x^(1/2)) ) / (1/(2(y^(1/2)))
take the derivative of that again^
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dy/dx = -y^(1/2) / x^(1/2)
Next use the quotient rule. (VdU-UdV)/V^2
-x^(1/2)/(2(y^1/2)) * dy/dx + (minus a negative) y^(1/2)/(2(x^(1/2))
all divided by x <-- (x^(1/2))^2
Now plug in dy/dx and that's your answer.
Interestingly, both y^1/2 and x^1/2 in the left term of the numerator cancel with the first derivative's terms, as well as 2/2, leaving d^2y/dx^2 = (1/2 + y^(1/2) / (2*x^(1/2))) / x