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

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

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

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

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

Solution steps

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

Treat

x, y

as constants

Apply the chain rule:

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

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

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

= \frac{1}{x + 2 y + 3 z} \cdot 3

Simplify

\frac{1}{x + 2 y + 3 z} \cdot 3 : \frac{3}{x + 2 y + 3 z}

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

x \to 3 lim (\frac{x ^{2} + x}{x - 3})

\int_{0}^{k} \frac{1}{2} (e^{x} - 1) d x

\int x csc^{2} (3 x) d x

\frac{\partial}{\partial y} (tan (7 + 2 x^{2} y^{3} z^{2}))

\frac{d}{d x} (cosh (2 x))

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