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

The answer to (\partial)/(\partial x)(-2xyz^2) is -2yz^2

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

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

- 2 y z^{2}

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the common derivative:

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

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

Simplify

= - 2 y z^{2}

\int_{0}^{1} \frac{1}{x ^{1}} d x

\int \frac{x ^{3} + 8}{( x ^{2} - 1 ) ( x - 2 )} d x

\frac{d y}{d x} + (\frac{3}{x} - 1) y = - \frac{2}{x}

y = 2 x, x = 3

\int e^{- s t} e^{a t} d t

What is (\partial)/(\partial x)(-2xyz^2) ?
The answer to (\partial)/(\partial x)(-2xyz^2) is -2yz^2