What is (\partial)/(\partial x)(7y-y^3-4x^2y) ?

The answer to (\partial)/(\partial x)(7y-y^3-4x^2y) is -8yx

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(\partial)/(\partial x)(7y-y^3-4x^2y)

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

- 8 y x

Solution steps

\frac{\partial}{\partial x} (7 y - y^{3} - 4 x^{2} y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (7 y) - \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (4 x^{2} y)

\frac{\partial}{\partial x} (7 y) = 0

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

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

= 0 - 0 - 8 y x

Simplify

= - 8 y x

y^{5} y^{^{'}} = - 2 cos (π x)

\int ln (3 x) d x

δ (t + 1)

\int \frac{1}{x 81 x ^{2} + 1} d x

\frac{1}{x ^{5}}

What is (\partial)/(\partial x)(7y-y^3-4x^2y) ?
The answer to (\partial)/(\partial x)(7y-y^3-4x^2y) is -8yx