Find the derivative of $$y^2=x$$ as a function of y.
i have found for the function of x, it will be $$\pm\frac{1}{2\sqrt x}$$
however for the function of y will be $$\frac{dy}{dx}=2y$$ ?
it looks too simple, and I'm sure it's wrong.
2 Answers
$\begingroup$$$ y^2 = x , 2 \, y \, y^{'} = 1, $$
$$ y' = \dfrac{dy}{dx} = \dfrac{1}{2y}. $$
$\endgroup$ $\begingroup$As a function fo $y$, you should have $\frac{dx}{dy}=\frac{1}{2y}$ which is the same as $\frac{1}{2\sqrt{x}}$. Note however that square is a function only if we choose the positive square root. So there is no need for the plus/or minus.
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