What is (\partial)/(\partial x)(ln(x+4y+7z)) ?

The answer to (\partial)/(\partial x)(ln(x+4y+7z)) is 1/(x+4y+7z)

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(\partial)/(\partial x)(ln(x+4y+7z))

\frac{\partial}{\partial x} (ln (x + 4 y + 7 z))

\frac{1}{x + 4 y + 7 z}

Solution steps

\frac{\partial}{\partial x} (ln (x + 4 y + 7 z))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{x + 4 y + 7 z} \frac{\partial}{\partial x} (x + 4 y + 7 z)

= \frac{1}{x + 4 y + 7 z} \frac{\partial}{\partial x} (x + 4 y + 7 z)

\frac{\partial}{\partial x} (x + 4 y + 7 z) = 1

= \frac{1}{x + 4 y + 7 z} \cdot 1

Simplify

= \frac{1}{x + 4 y + 7 z}

\frac{d}{d x} ((x - 1)^{2} + 1)

\frac{\partial}{\partial x} (cos (y e^{x \cdot y}))

\int [1 + \frac{1}{( x + 1 ) ^{2}}] d x

x \to 2 lim (4 - x)

y = \frac{x ^{5}}{3 - x ^{4}}

What is (\partial)/(\partial x)(ln(x+4y+7z)) ?
The answer to (\partial)/(\partial x)(ln(x+4y+7z)) is 1/(x+4y+7z)