What is (\partial)/(\partial y)(54-2/3 x^2-4y^2) ?

The answer to (\partial)/(\partial y)(54-2/3 x^2-4y^2) is -8y

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(\partial)/(\partial y)(54-2/3 x^2-4y^2)

\frac{\partial}{\partial y} (54 - \frac{2}{3} x^{2} - 4 y^{2})

- 8 y

Solution steps

\frac{\partial}{\partial y} (54 - \frac{2}{3} x^{2} - 4 y^{2})

Simplify

54 - \frac{2}{3} x^{2} - 4 y^{2} : 54 - \frac{2 x ^{2}}{3} - 4 y^{2}

= \frac{\partial}{\partial y} (54 - \frac{2 x ^{2}}{3} - 4 y^{2})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (54) - \frac{\partial}{\partial y} (\frac{2 x ^{2}}{3}) - \frac{\partial}{\partial y} (4 y^{2})

\frac{\partial}{\partial y} (54) = 0

\frac{\partial}{\partial y} (\frac{2 x ^{2}}{3}) = 0

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

= 0 - 0 - 8 y

Simplify

= - 8 y

n = 1 \sum \infty \frac{n}{e ^{n^{3}}}

cos (- 4 t)

\frac{d}{d x} (sin (x) + \frac{1}{2} cot (x))

y = x + ln (3 x)

\int_{0}^{\frac{3 π}{4}} sec^{4} (\frac{x}{3}) d x

What is (\partial)/(\partial y)(54-2/3 x^2-4y^2) ?
The answer to (\partial)/(\partial y)(54-2/3 x^2-4y^2) is -8y