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

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

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

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

- 8 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 8 y

Simplify

= - 8 y

(0, - 2), (- \frac{1}{2}, 3)

n = 5 \sum \infty \frac{1}{n ( n + 1 )}

\frac{d y}{d x} = sin (5 x)

\int 5 x - 1 d x

\frac{d P}{d t} + 2 tP = P + 8 t - 4

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