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

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

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

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

- x z^{2}

Solution steps

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

Treat

x, z

as constants

Take the constant out:

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

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

Apply the common derivative:

\frac{\partial}{\partial y} (y) = 1

= - x z^{2} \cdot 1

Simplify

= - x z^{2}

y = x^{2} 25 - x^{2}

\int \frac{e ^{x}}{6 + e ^{x}} d x

\int_{- 1}^{1} e^{y} - (y^{2} - 5) d y

\int_{1}^{5} (2 x - 1) d x

\int \frac{3 x}{2} d x

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