What is (\partial)/(\partial y)(2x^3+4y-2z^2) ?

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

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(\partial)/(\partial y)(2x^3+4y-2z^2)

\frac{\partial}{\partial y} (2 x^{3} + 4 y - 2 z^{2})

4

Solution steps

\frac{\partial}{\partial y} (2 x^{3} + 4 y - 2 z^{2})

Treat

x, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (2 x^{3}) + \frac{\partial}{\partial y} (4 y) - \frac{\partial}{\partial y} (2 z^{2})

\frac{\partial}{\partial y} (2 x^{3}) = 0

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

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

= 0 + 4 - 0

Simplify

= 4

\frac{d}{d x} (\frac{x ^{2} - 3 x - 54}{x + 6})

f (x) = 5 - 2 x

y = x + 2, y = x^{2}, y = 0

\int sin^{2} (k x) d x

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