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

The answer to (\partial)/(\partial x)(3y^3-4yx^2) is -8yx

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

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

- 8 y x

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (4 y x^{2}) = 8 y x

= 0 - 8 y x

Simplify

= - 8 y x

\int_{0}^{7} f (x) d x

θ \to 0 lim (\frac{sin ( 4 θ )}{θ})

f (x) = 8 x^{2} + 4 e^{- 0.5 x}

3 x

(x - y) d x + d y = 0

What is (\partial)/(\partial x)(3y^3-4yx^2) ?
The answer to (\partial)/(\partial x)(3y^3-4yx^2) is -8yx