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

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

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(\partial)/(\partial y)(x^2-y^2)

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

- 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \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

= 0 - 2 y

Simplify

= - 2 y

\int x^{2} + 4 x + 3 d x

\int \frac{sec ( x )}{1 - tan ( x )} d x

\frac{d}{d x} ((x^{3} - 7) (2 x^{2} + 3))

(2 x + 4)^{^{'}}

\int_{e}^{4 e} \frac{1}{x} d x

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