What is (\partial)/(\partial x)(ln(x+3y+7z)) ?

The answer to (\partial)/(\partial x)(ln(x+3y+7z)) is 1/(x+3y+7z)

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(\partial)/(\partial x)(ln(x+3y+7z))

\frac{\partial}{\partial x} (ln (x + 3 y + 7 z))

\frac{1}{x + 3 y + 7 z}

Solution steps

\frac{\partial}{\partial x} (ln (x + 3 y + 7 z))

Treat

y, z

as constants

Apply the chain rule:

\frac{1}{x + 3 y + 7 z} \frac{\partial}{\partial x} (x + 3 y + 7 z)

= \frac{1}{x + 3 y + 7 z} \frac{\partial}{\partial x} (x + 3 y + 7 z)

\frac{\partial}{\partial x} (x + 3 y + 7 z) = 1

= \frac{1}{x + 3 y + 7 z} \cdot 1

Simplify

= \frac{1}{x + 3 y + 7 z}

\int tan^{3} (x) (sec (x))^{\frac{3}{2}} d x

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

x \to 0 lim (\frac{3}{0})

n = 0 \sum \infty (\frac{1}{3})^{2 n}

\frac{5 x}{4 x ^{2} + 1}

What is (\partial)/(\partial x)(ln(x+3y+7z)) ?
The answer to (\partial)/(\partial x)(ln(x+3y+7z)) is 1/(x+3y+7z)