What is (\partial)/(\partial x)(ln((2xy)/(7z))) ?

The answer to (\partial)/(\partial x)(ln((2xy)/(7z))) is 1/x

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(\partial)/(\partial x)(ln((2xy)/(7z)))

\frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

\frac{1}{x}

Solution steps

\frac{\partial}{\partial x} (ln (\frac{2 x y}{7 z}))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{\frac{2 x y}{7 z}} \frac{\partial}{\partial x} (\frac{2 x y}{7 z})

= \frac{1}{\frac{2 x y}{7 z}} \frac{\partial}{\partial x} (\frac{2 x y}{7 z})

\frac{\partial}{\partial x} (\frac{2 x y}{7 z}) = \frac{2 y}{7 z}

= \frac{1}{\frac{2 x y}{7 z}} \cdot \frac{2 y}{7 z}

Simplify

\frac{1}{\frac{2 x y}{7 z}} \cdot \frac{2 y}{7 z} : \frac{1}{x}

= \frac{1}{x}

x \to \infty lim (\frac{x}{5})

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

\int_{- 1}^{2} (20 - x^{4}) d x

\frac{d}{d x} (8 x + 4)

What is (\partial)/(\partial x)(ln((2xy)/(7z))) ?
The answer to (\partial)/(\partial x)(ln((2xy)/(7z))) is 1/x