What is (\partial}{\partial y}(-\frac{y^3)/3-8x) ?

The answer to (\partial}{\partial y}(-\frac{y^3)/3-8x) is -y^2

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(\partial}{\partial y}(-\frac{y^3)/3-8x)

\frac{\partial}{\partial y} (- \frac{y ^{3}}{3} - 8 x)

- y^{2}

Solution steps

\frac{\partial}{\partial y} (- \frac{y ^{3}}{3} - 8 x)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial y} (\frac{y ^{3}}{3}) - \frac{\partial}{\partial y} (8 x)

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

\frac{\partial}{\partial y} (8 x) = 0

= - y^{2} - 0

Simplify

= - y^{2}

n \to \infty lim (\frac{n}{10 + n})

\frac{( s + 2 ) ^{2}}{s ^{3}}

x \to 3 lim (\frac{x - 3}{x ^{2} + 9})

x \to 0 lim (x - 2)

\int \frac{x + 5}{x ^{2} + 4 x + 3} d x

What is (\partial}{\partial y}(-\frac{y^3)/3-8x) ?
The answer to (\partial}{\partial y}(-\frac{y^3)/3-8x) is -y^2