What is (\partial)/(\partial x)(-8xy-2x^4-2y^4) ?

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

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(\partial)/(\partial x)(-8xy-2x^4-2y^4)

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

- 8 x^{3} - 8 y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (8 x y) = 8 y

\frac{\partial}{\partial x} (2 x^{4}) = 8 x^{3}

\frac{\partial}{\partial x} (2 y^{4}) = 0

= - 8 y - 8 x^{3} - 0

Simplify

= - 8 x^{3} - 8 y

x \to 0 lim (\frac{6 ^{x} - 5 ^{x}}{x})

f (x) = \frac{x - 3}{x + 3}

\frac{\partial}{\partial y} (\frac{x}{5 + y})

x \to - 1 lim (x - 1)

\frac{d ^{2}}{d t ^{2}} x + 5 \frac{d}{d t} x + 2 x = 0

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