What is (\partial)/(\partial y)(2x^2-x^4+y^2) ?

The answer to (\partial)/(\partial y)(2x^2-x^4+y^2) is 2y

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(\partial)/(\partial y)(2x^2-x^4+y^2)

\frac{\partial}{\partial y} (2 x^{2} - x^{4} + y^{2})

2 y

Solution steps

\frac{\partial}{\partial y} (2 x^{2} - x^{4} + y^{2})

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} (x^{4}) + \frac{\partial}{\partial y} (y^{2})

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

\frac{\partial}{\partial y} (x^{4}) = 0

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

= 0 - 0 + 2 y

Simplify

= 2 y

\frac{\partial}{\partial x} (2 x^{2} + y^{2} + 2 x y + y x + 7 y)

\frac{\partial}{\partial v} (ln (u cos (v)))

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

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

x \to \infty lim (- 1)

What is (\partial)/(\partial y)(2x^2-x^4+y^2) ?
The answer to (\partial)/(\partial y)(2x^2-x^4+y^2) is 2y