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

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

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

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

- 2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 2 y

Simplify

= - 2 y

- \frac{5}{4 y ^{3}}

\frac{d}{d x} (cos (x^{- 5}))

\int (2 sin (θ) - csc (θ))^{2} d θ

y = (4 x - 5) 2

(x + \frac{9}{x})^{^{'}}

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