What is (\partial)/(\partial x)(x^2yz^2+sin(yz)) ?

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

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(\partial)/(\partial x)(x^2yz^2+sin(yz))

\frac{\partial}{\partial x} (x^{2} y z^{2} + sin (yz))

2 x y z^{2}

Solution steps

\frac{\partial}{\partial x} (x^{2} y z^{2} + sin (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} (sin (yz))

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

\frac{\partial}{\partial x} (sin (yz)) = 0

= 2 y z^{2} x + 0

Simplify

= 2 x y z^{2}

\frac{d}{d x} (2 + sin (3 x))

\int sin (x^{2} + y^{2})

A^{^{'}}

\int 2 sin (2 x) cos (2 x) d x

\int \frac{( x - 1 )}{( x + 4 )} d x

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