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

The answer to (\partial)/(\partial y)(4xy-2x^2-y^4) is 4x-4y^3

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

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

4 x - 4 y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (y^{4}) = 4 y^{3}

= 4 x - 0 - 4 y^{3}

Simplify

= 4 x - 4 y^{3}

- 8 sin (x)

\frac{d}{d x} (x^{4} + 2 x^{2} + 1)

\frac{d}{d t} (\frac{cos ( t )}{t})

\int_{0}^{1} x^{2} e^{- x} d x

\int (x - 4)^{3} d x

What is (\partial)/(\partial y)(4xy-2x^2-y^4) ?
The answer to (\partial)/(\partial y)(4xy-2x^2-y^4) is 4x-4y^3