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

The answer to (\partial)/(\partial x)(2z^3-3(x^2+y^2)z) is -6zx

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

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

- 6 z x

Solution steps

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

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (3 (x^{2} + y^{2}) z) = 6 z x

= 0 - 6 z x

Simplify

= - 6 z x

x \to - 8 + lim (\frac{x}{x + 8})

\frac{\partial}{\partial x} (6 x^{3} y + 6 y^{5} + 1)

\frac{d}{d v} (v - 1)

\int_{1}^{R} x^{- \frac{7}{5}} d x

\frac{d}{d x} ((6 x^{5} - 4 x^{4} + 7 x) (x^{3} - x^{2} - 1))

What is (\partial)/(\partial x)(2z^3-3(x^2+y^2)z) ?
The answer to (\partial)/(\partial x)(2z^3-3(x^2+y^2)z) is -6zx