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

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

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

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (4 x^{2}) = 0

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

= 0 + 2 y

Simplify

= 2 y

\frac{\partial}{\partial y} (8 - 2 x^{2} + y^{2})

f (x) = x^{2} (x^{2} + 1)

\int e^{6 x + 9} d x

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

\int_{0}^{4} x 16 - x^{2} d x

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