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

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

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

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

2 y

Solution steps

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

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

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

= 0 + 2 y

Simplify

= 2 y

x^{2} \frac{d y}{d x} = x y + x^{2} e^{\frac{y}{x}}

y^{^{''}} + 6 y + 8 = 0, y (0) = 0, y^{^{'}} (0) = 12

f (x) = 4 x^{6} - 5 x^{- 5} + 3 x - \frac{6}{x ^{4}} - 5 e^{x}

\frac{d}{d x} (cos (3 x + 2) - (2 x + 1)^{- 3})

\int \frac{sin ( y )}{x} d x

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