# How do you find (df)/dy and (df)/dx of f(x,y)=(3x^2-2e^-y)/(2e^x+y^-3), using the quotient rule?

Feb 11, 2017

$\frac{\partial f}{\partial x} = \frac{6 x \left(2 {e}^{x} + {y}^{-} 3\right) - 2 {e}^{x} \left(3 {x}^{2} - 2 {e}^{-} y\right)}{2 {e}^{x} + {y}^{-} 3} ^ 2$

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

#### Explanation:

The quotient rule for multi variable calculus is identical to single variable calculus except that we perform the differentiation wrt to a specific variable whilst holding other variable constant. so:

$\frac{\partial}{\partial x} \frac{u \left(x , y\right)}{v \left(x , y\right)} = \frac{v \frac{\partial u}{\partial x} - u \frac{\partial u}{\partial x}}{v} ^ 2$

$\frac{\partial}{\partial y} \frac{u \left(x , y\right)}{v \left(x , y\right)} = \frac{v \frac{\partial u}{\partial y} - u \frac{\partial u}{\partial y}}{v} ^ 2$

We have:

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

If we differentiate wrt $x$ using the quotient rule (whilst holding $y$ constant) we get:

$\frac{\partial f}{\partial x} = \frac{\left(2 {e}^{x} + {y}^{-} 3\right) \left(6 x\right) - \left(3 {x}^{2} - 2 {e}^{-} y\right) \left(2 {e}^{x}\right)}{2 {e}^{x} + {y}^{-} 3} ^ 2$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{6 x \left(2 {e}^{x} + {y}^{-} 3\right) - 2 {e}^{x} \left(3 {x}^{2} - 2 {e}^{-} y\right)}{2 {e}^{x} + {y}^{-} 3} ^ 2$

If we differentiate wrt $y$ using the quotient rule (whilst holding $x$ constant) we get:

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