What is (\partial)/(\partial y)(z^3-xy+yz+y^3-2) ?

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

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(\partial)/(\partial y)(z^3-xy+yz+y^3-2)

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

z + 3 y^{2} - x

Solution steps

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

Treat

z, x

as constants

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial y} (yz) = z

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

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

= 0 - x + z + 3 y^{2} - 0

Simplify

= z + 3 y^{2} - x

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\frac{d}{d x} (tan^{- 1} (x))

2 x, x^{2}, 0, 2

\int \frac{1}{2} e^{- 2 x} d x

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