What is (\partial)/(\partial x)(-xyz^2+16) ?

The answer to (\partial)/(\partial x)(-xyz^2+16) is -yz^2

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(\partial)/(\partial x)(-xyz^2+16)

\frac{\partial}{\partial x} (- x y z^{2} + 16)

- y z^{2}

Solution steps

\frac{\partial}{\partial x} (- x y z^{2} + 16)

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= - \frac{\partial}{\partial x} (x y z^{2}) + \frac{\partial}{\partial x} (16)

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

\frac{\partial}{\partial x} (16) = 0

= - y z^{2} + 0

Simplify

= - y z^{2}

y = x - x^{3} (1)

\frac{5}{s ^{2} + 4}

\frac{\partial}{\partial x} (- 4 x^{5} y^{- 5})

\frac{\partial}{\partial y} (1 + x y^{2})

\frac{d}{d x} (ln (x^{11}))

What is (\partial)/(\partial x)(-xyz^2+16) ?
The answer to (\partial)/(\partial x)(-xyz^2+16) is -yz^2