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

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

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

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

- 4 y

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= - 2 \cdot 2 y^{2 - 1}

Simplify

= - 4 y

\int_{- 2}^{3} \frac{39}{x ^{4}} d x

f (θ) = cos (θ^{2})

\frac{\partial}{\partial x} (ln (y^{2}))

y^{^{''}} + b y (x) = 0

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