What is (\partial}{\partial y}(\frac{x^2-y^2)/2) ?

The answer to (\partial}{\partial y}(\frac{x^2-y^2)/2) is -y

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(\partial}{\partial y}(\frac{x^2-y^2)/2)

\frac{\partial}{\partial y} (\frac{x ^{2} - y ^{2}}{2})

- y

Solution steps

\frac{\partial}{\partial y} (\frac{x ^{2} - y ^{2}}{2})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{2} \frac{\partial}{\partial y} (x^{2} - y^{2})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{1}{2} (\frac{\partial}{\partial y} (x^{2}) - \frac{\partial}{\partial y} (y^{2}))

\frac{\partial}{\partial y} (x^{2}) = 0

\frac{\partial}{\partial y} (y^{2}) = 2 y

= \frac{1}{2} (0 - 2 y)

Simplify

\frac{1}{2} (0 - 2 y) : - y

= - y

\int_{0}^{0.02} 200 x d x

\frac{3 s + 1}{s ( s + 2 )}

\frac{1}{2 x + 2}, at x = 5

(4 - x^{2}) \frac{d y}{d x} = 2 y

y = (\frac{x + 1}{x - 3})^{2}, (7, 4)

What is (\partial}{\partial y}(\frac{x^2-y^2)/2) ?
The answer to (\partial}{\partial y}(\frac{x^2-y^2)/2) is -y