What is (\partial)/(\partial y)(ln(x-2y+3z)) ?

The answer to (\partial)/(\partial y)(ln(x-2y+3z)) is -2/(x-2y+3z)

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(\partial)/(\partial y)(ln(x-2y+3z))

\frac{\partial}{\partial y} (ln (x - 2 y + 3 z))

- \frac{2}{x - 2 y + 3 z}

Solution steps

\frac{\partial}{\partial y} (ln (x - 2 y + 3 z))

Treat

x, z

as constants

Apply the chain rule:

\frac{1}{x - 2 y + 3 z} \frac{\partial}{\partial y} (x - 2 y + 3 z)

= \frac{1}{x - 2 y + 3 z} \frac{\partial}{\partial y} (x - 2 y + 3 z)

\frac{\partial}{\partial y} (x - 2 y + 3 z) = - 2

= \frac{1}{x - 2 y + 3 z} (- 2)

Simplify

\frac{1}{x - 2 y + 3 z} (- 2) : - \frac{2}{x - 2 y + 3 z}

= - \frac{2}{x - 2 y + 3 z}

What is (\partial)/(\partial y)(ln(x-2y+3z)) ?
The answer to (\partial)/(\partial y)(ln(x-2y+3z)) is -2/(x-2y+3z)