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

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

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

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

\frac{1}{x + y}

Solution steps

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

Treat

x

as a constant

Apply the chain rule:

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

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

\frac{\partial}{\partial y} (∣ x + y ∣) = \frac{x + y}{∣ x + y ∣}

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

Simplify

\frac{1}{∣ x + y ∣} \cdot \frac{x + y}{∣ x + y ∣} : \frac{1}{x + y}

= \frac{1}{x + y}

\frac{d y}{d x} = e^{3 y + 10 x}

\frac{d}{d x} (sin (x) (1 - cos (x)))

n = 0 \sum \infty \frac{1}{2 - e ^{- n}}

f (x) = - 10 x sin (5 x^{2} + 5)

\frac{d}{d x} ((4 - x^{\frac{2}{3}})^{\frac{2}{3}})

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