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

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

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

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

\frac{1}{3 x + y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

\int (x - 1)^{2} x d x

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

x \to 0 lim (\frac{4}{x} - \frac{4}{e ^{x} - 1})

\frac{\partial}{\partial x} ((6 x + 3 y) ey)

\int \frac{1}{( x - 3 ) ( x + 2 )} d x

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