What is the partial derivative with respect to y of squareRoot(x-y)? Thanks.
$\displaystyle \displaystyle f = \sqrt{x-y}$
$\displaystyle \displaystyle f = (x-y)^\frac{1}{2}$
Hold $\displaystyle x$ as constant
$\displaystyle \displaystyle \frac{\partial f}{\partial y} = \frac{1}{2}(-1)(x-y)^\frac{-1}{2}$
$\displaystyle \displaystyle \frac{\partial f}{\partial y} = \frac{-1}{2\sqrt{x-y}}$