What is (\partial)/(\partial x)(4xyz+ln(2xyz)) ?

The answer to (\partial)/(\partial x)(4xyz+ln(2xyz)) is 4yz+1/x

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(\partial)/(\partial x)(4xyz+ln(2xyz))

\frac{\partial}{\partial x} (4 x yz + ln (2 x yz))

4 yz + \frac{1}{x}

Solution steps

\frac{\partial}{\partial x} (4 x yz + ln (2 x yz))

Treat

y, z

as constants

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (4 x yz) + \frac{\partial}{\partial x} (ln (2 x yz))

\frac{\partial}{\partial x} (4 x yz) = 4 yz

\frac{\partial}{\partial x} (ln (2 x yz)) = \frac{1}{x}

= 4 yz + \frac{1}{x}

\frac{2 x ^{4} + 1}{y} y^{^{'}} + \frac{1}{6} y^{3} x^{3} = 0

y^{^{'}} - y^{2} = 36

\int_{0}^{1} \frac{30}{4 y - 1} d y

x \cdot \frac{d y}{d x} = 4 y

f (e^{2 x})

What is (\partial)/(\partial x)(4xyz+ln(2xyz)) ?
The answer to (\partial)/(\partial x)(4xyz+ln(2xyz)) is 4yz+1/x