What is (\partial)/(\partial y)(4y^3-4x) ?

The answer to (\partial)/(\partial y)(4y^3-4x) is 12y^2

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(\partial)/(\partial y)(4y^3-4x)

\frac{\partial}{\partial y} (4 y^{3} - 4 x)

12 y^{2}

Solution steps

\frac{\partial}{\partial y} (4 y^{3} - 4 x)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial y} (4 y^{3}) = 12 y^{2}

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

= 12 y^{2} - 0

Simplify

= 12 y^{2}

23 cos (x^{2})

\int x^{2} (2 x^{3} + 1)^{7} d x

e^{x} 5

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

\int_{- 1}^{1} \frac{1}{x ^{2} - x - 2} d x

What is (\partial)/(\partial y)(4y^3-4x) ?
The answer to (\partial)/(\partial y)(4y^3-4x) is 12y^2