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

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

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

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

\frac{1}{x + y}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

= \frac{1}{x + y}

f (u) = u^{2}

\frac{1}{3 + x}

\frac{x}{x ^{2} - 1}

\int 6 e^{3 x} - \frac{8}{e ^{2 x}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) e^{2 x}

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