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

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

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

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

- 4 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

= 0 - 4 y

Simplify

= - 4 y

x \to \infty lim (\frac{x}{( ln ( x ) ) ^{x}})

\frac{\partial}{\partial x} (n^{2} x \cdot (1 - x)^{n})

\int \frac{1}{( x ^{2} + 2 x + 2 ) ^{2}} d x

\frac{d}{d t} (2 sin (4 t))

\frac{d y}{d t} + 2 y = 1

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