How do you find (delf(x,y))/(dely) and (delf(x,y))/(delx) of f(x,y)=(yx^2-2y)/(2ye^x+y^-3), using the quotient rule?

Jun 19, 2018

$\frac{\partial f \left(x , y\right)}{\partial y} = \frac{\left(2 y {e}^{x} + {y}^{-} 3\right) \left({x}^{2} - 2\right) - \left(y {x}^{2} - 2 y\right) \left(2 {e}^{x} - 3 {y}^{-} 4\right)}{2 y {e}^{x} + {y}^{-} 3} ^ 2$

$\frac{\partial f \left(x , y\right)}{\partial x} = \frac{2 y x \left(2 y {e}^{x} + {y}^{-} 3\right) - 2 y {e}^{x} \left(y {x}^{2} - 2 y\right)}{2 y {e}^{x} + {y}^{-} 3} ^ 2$

Explanation:

To partially differentiate, we treat one variable as a constant.

So, for $\frac{\partial f \left(x , y\right)}{\partial y}$ we treat $x$ as a constant.

$\frac{\partial}{\partial x} \frac{u}{v} = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial v}{\partial x}}{v} ^ 2$
$\frac{\partial}{\partial y} \frac{u}{v} = \frac{v \frac{\partial u}{\partial y} - u \frac{\partial v}{\partial y}}{v} ^ 2$

$u = y {x}^{2} - 2 y$
$\frac{\partial u}{\partial y} = {x}^{2} - 2$

$v = 2 y {e}^{x} + {y}^{-} 3$
$\frac{\partial v}{\partial y} = 2 {e}^{x} - 3 {y}^{-} 4$

$\frac{\partial f \left(x , y\right)}{\partial y} = \frac{\left(2 y {e}^{x} + {y}^{-} 3\right) \left({x}^{2} - 2\right) - \left(y {x}^{2} - 2 y\right) \left(2 {e}^{x} - 3 {y}^{-} 4\right)}{2 y {e}^{x} + {y}^{-} 3} ^ 2$

Now, for $\frac{\partial f \left(x , y\right)}{\partial x}$ we treat $y$ as a constant.

$u = y {x}^{2} - 2 y$
$\frac{\partial u}{\partial x} = 2 y x$

$v = 2 y {e}^{x} + {y}^{-} 3$
$\frac{\partial v}{\partial x} = 2 y {e}^{x}$

$\frac{\partial f \left(x , y\right)}{\partial y} = \frac{2 y x \left(2 y {e}^{x} + {y}^{-} 3\right) - 2 y {e}^{x} \left(y {x}^{2} - 2 y\right)}{2 y {e}^{x} + {y}^{-} 3} ^ 2$