What is (\partial)/(\partial x)(ln((xy)/(y+z))) ?

The answer to (\partial)/(\partial x)(ln((xy)/(y+z))) is 1/x

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(\partial)/(\partial x)(ln((xy)/(y+z)))

\frac{\partial}{\partial x} (ln (\frac{x y}{y + z}))

\frac{1}{x}

Solution steps

\frac{\partial}{\partial x} (ln (\frac{x y}{y + z}))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{\frac{x y}{y + z}} \frac{\partial}{\partial x} (\frac{x y}{y + z})

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

\frac{\partial}{\partial x} (\frac{x y}{y + z}) = \frac{y}{y + z}

= \frac{1}{\frac{x y}{y + z}} \cdot \frac{y}{y + z}

Simplify

\frac{1}{\frac{x y}{y + z}} \cdot \frac{y}{y + z} : \frac{1}{x}

= \frac{1}{x}

\frac{6}{s ( s ^{2} + 2 s + 6 )}

f (x) = \frac{1 + x ^{2}}{x ^{2} - 1}

y = ln (x + x^{2} - 6)

\frac{d}{d a} (csc (a) tan (a) cos (a) - csc^{2} (a))

x \to 0 lim (ln (x^{2} + 1))

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