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

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

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

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

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

Solution steps

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

Treat

z, y

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x, g = ln (x + 2 y + 3 z)

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

\frac{\partial}{\partial x} (x) = 1

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

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

Simplify

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

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

\frac{d}{d x} (e^{- 5})

\frac{\partial}{\partial y} (e^{x} sin (yz))

\frac{d}{d x} (5 x^{2} - 3 x)

x \to - 1 + lim (\frac{x - 1}{x + 1})

- \frac{192}{x ^{4}}

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