What is (\partial)/(\partial x)(xyzln(x+y)) ?

The answer to (\partial)/(\partial x)(xyzln(x+y)) is yz(ln(x+y)+x/(x+y))

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(\partial)/(\partial x)(xyzln(x+y))

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

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

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the Product Rule:

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

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

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

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

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

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

Simplify

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

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

x \to 4 - lim (8 - 2 x)

x \to - 8 - lim (\frac{- 2 x}{x + 8})

\int π (x + 2)^{2} d x

f (x) = x + 63, (1, 8)

\int_{0}^{1} t 1 + 4 t^{2} d t

What is (\partial)/(\partial x)(xyzln(x+y)) ?
The answer to (\partial)/(\partial x)(xyzln(x+y)) is yz(ln(x+y)+x/(x+y))