What is (\partial)/(\partial x)(x^2yz^2-4cos(3yz)) ?

The answer to (\partial)/(\partial x)(x^2yz^2-4cos(3yz)) is 2xyz^2

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(\partial)/(\partial x)(x^2yz^2-4cos(3yz))

\frac{\partial}{\partial x} (x^{2} y z^{2} - 4 cos (3 yz))

2 x y z^{2}

Solution steps

\frac{\partial}{\partial x} (x^{2} y z^{2} - 4 cos (3 yz))

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2} y z^{2}) - \frac{\partial}{\partial x} (4 cos (3 yz))

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

\frac{\partial}{\partial x} (4 cos (3 yz)) = 0

= 2 y z^{2} x - 0

Simplify

= 2 x y z^{2}

f (x) = - 3 x^{2} + 1 x, at x = 0

\frac{d}{d x} (1 + 3 sin^{3} (x))

x^{2} + 1 + x^{2}

\int_{\frac{π}{4}}^{x} sec^{2} (t) d t

\int \frac{x + 1}{x ^{2} - 2 x - 1} d x

What is (\partial)/(\partial x)(x^2yz^2-4cos(3yz)) ?
The answer to (\partial)/(\partial x)(x^2yz^2-4cos(3yz)) is 2xyz^2