The discussion focuses on proving that the second derivative of the equation \(\sqrt{x} + \sqrt{y} = 4\) is \(y'' = \frac{2}{x\sqrt{x}}\). The initial attempt at finding the first derivative, \(\frac{dy}{dx} = -\frac{\sqrt{y}}{\sqrt{x}}\), is confirmed to be correct. However, the calculation of the second derivative contains errors, particularly in the application of the quotient rule. Participants in the discussion seek clarification on the correct method to derive the second derivative accurately.
PREREQUISITESStudents studying calculus, particularly those focusing on implicit differentiation and second derivatives, as well as educators looking for examples of common errors in derivative calculations.
Sakha said: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}
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Which is wrong if I plug in values.
Anyone sees my error?