$\displaystyle 2x^2y^(\frac{1}{2})$
Is $\displaystyle f_y = 2x^2\frac{1}{2}y^{-\frac{1}{2}}$?
If im wrong can you please explain why and how to find it properly.
Thanks.
Are you asking to find $\displaystyle \frac{\partial f}{\partial y}$ if $\displaystyle f(x, y) = 2x^2y^{\frac{1}{2}}$?
If so, leave $\displaystyle x$ constant and differentiate with respect to $\displaystyle y$.
So $\displaystyle \frac{\partial f}{\partial y} = \frac{1}{2}\cdot 2x^2 y^{-\frac{1}{2}}$
$\displaystyle = \frac{x^2}{\sqrt{y}}$
$\displaystyle = \frac{x^2\sqrt{y}}{y}$.
If x is treated as a constant, so is $\displaystyle x^2$.
Maybe I should write it out like this
$\displaystyle f(x, y) = 2x^2 y^{\frac{1}{2}}$
$\displaystyle \frac{\partial f}{\partial y} = 2x^2 \frac{\partial}{\partial y} \left(y^{\frac{1}{2}}\right)$, since $\displaystyle 2x^2$ is a constant.
$\displaystyle = 2x^2 \cdot \frac{1}{2}y^{-\frac{1}{2}}$
$\displaystyle = x^2 y^{-\frac{1}{2}}$.