What is (\partial)/(\partial y)(2x^2-y^2+3x-y) ?

The answer to (\partial)/(\partial y)(2x^2-y^2+3x-y) is -2y-1

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(\partial)/(\partial y)(2x^2-y^2+3x-y)

\frac{\partial}{\partial y} (2 x^{2} - y^{2} + 3 x - y)

- 2 y - 1

Solution steps

\frac{\partial}{\partial y} (2 x^{2} - y^{2} + 3 x - y)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (2 x^{2}) - \frac{\partial}{\partial y} (y^{2}) + \frac{\partial}{\partial y} (3 x) - \frac{\partial}{\partial y} (y)

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

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

\frac{\partial}{\partial y} (3 x) = 0

\frac{\partial}{\partial y} (y) = 1

= 0 - 2 y + 0 - 1

Simplify

= - 2 y - 1

ln (cos (2 x))

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What is (\partial)/(\partial y)(2x^2-y^2+3x-y) ?
The answer to (\partial)/(\partial y)(2x^2-y^2+3x-y) is -2y-1