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}

\int_{- e^{2}}^{- e} \frac{3}{x} d x

\int \frac{1}{x ^{2} + 4 x + 3} d x

sin (x y)

f (x) = (x^{3} - 4 x)^{8}, (2, 0)

\int (3 y - 2 y^{- 3}) d y

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