1. ## Partial derivative confused.

$2x^2y^(\frac{1}{2})$

Is $f_y = 2x^2\frac{1}{2}y^{-\frac{1}{2}}$?

If im wrong can you please explain why and how to find it properly.

Thanks.

2. Originally Posted by el123
$2x^2y^(\frac{1}{2})$

Is $f_y = 2x^2\frac{1}{2}y^{-\frac{1}{2}}$?

If im wrong can you please explain why and how to find it properly.

Thanks.
Are you asking to find $\frac{\partial f}{\partial y}$ if $f(x, y) = 2x^2y^{\frac{1}{2}}$?

If so, leave $x$ constant and differentiate with respect to $y$.

So $\frac{\partial f}{\partial y} = \frac{1}{2}\cdot 2x^2 y^{-\frac{1}{2}}$

$= \frac{x^2}{\sqrt{y}}$

$= \frac{x^2\sqrt{y}}{y}$.

3. yeah thats what i wanted to know but i'm confused why you don't bring the 2 down as well.

4. Originally Posted by el123
yeah thats what i wanted to know but i'm confused why you don't bring the 2 down as well.
If x is treated as a constant, so is $x^2$.

Maybe I should write it out like this

$f(x, y) = 2x^2 y^{\frac{1}{2}}$

$\frac{\partial f}{\partial y} = 2x^2 \frac{\partial}{\partial y} \left(y^{\frac{1}{2}}\right)$, since $2x^2$ is a constant.

$= 2x^2 \cdot \frac{1}{2}y^{-\frac{1}{2}}$

$= x^2 y^{-\frac{1}{2}}$.