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

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

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

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

4 x y

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply the Power Rule:

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

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

Simplify

= 4 x y

\int_{0}^{8 π} x^{2} sin (2 x) d x

sin (tan (ln (x^{2} + 5)))

f (x) = - 15 x^{2} - 14 x + 8

t^{3} e^{- 0.1 t}

\int_{1}^{\infty} \frac{1}{e ^{x}} d x

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