What is (\partial)/(\partial x)(-4xy-x^4-y^4) ?

The answer to (\partial)/(\partial x)(-4xy-x^4-y^4) is -4x^3-4y

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(\partial)/(\partial x)(-4xy-x^4-y^4)

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

- 4 x^{3} - 4 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

\frac{\partial}{\partial x} (y^{4}) = 0

= - 4 y - 4 x^{3} - 0

Simplify

= - 4 x^{3} - 4 y

f (x) = x + 4

\int \frac{x ^{2} - x + 14}{x ^{3} + 7 x} d x

\int \frac{x ^{2} + 2 x}{x ^{3} + 3 x ^{2} + 2} d x

y^{^{''}} - y = t^{2} e^{t}

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