What is (\partial)/(\partial x)(z+yzsin(2xyz)) ?

The answer to (\partial)/(\partial x)(z+yzsin(2xyz)) is 2z^2y^2cos(2zyx)

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(\partial)/(\partial x)(z+yzsin(2xyz))

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

2 z^{2} y^{2} cos (2 zy x)

Solution steps

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

Treat

z, y

as constants

Apply the Sum/Difference Rule:

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

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

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

\frac{\partial}{\partial x} (yz sin (2 x yz)) = 2 z^{2} y^{2} cos (2 zy x)

= 0 + 2 z^{2} y^{2} cos (2 zy x)

Simplify

= 2 z^{2} y^{2} cos (2 zy x)

\int_{- 4}^{- 2} (- x + ∣ - 3 x - 9 ∣) d x

\frac{d y}{d x} = 2 y - y^{2}

\frac{d}{d x} (2 x^{7} - 3 x^{6} + 3 x^{3} - 4 x^{2} - 7)

\int \frac{7 - 7 x}{1 - x} d x

\int \frac{7}{3 - 3 x ^{2}} d x

What is (\partial)/(\partial x)(z+yzsin(2xyz)) ?
The answer to (\partial)/(\partial x)(z+yzsin(2xyz)) is 2z^2y^2cos(2zyx)