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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Treat

z, y

as constants

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

\frac{4 d y}{d x} = y^{2}

arctan (x)

n = 1 \sum \infty \frac{n}{n}

f (x) = 4^{x} x^{4}

\frac{e ^{- 2 s}}{s - 2}

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