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

The answer to (\partial)/(\partial x)(ln((x+y)/(x-y))) is -(2y)/((x+y)(x-y))

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

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

- \frac{2 y}{( x + y ) ( x - y )}

Solution steps

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

Treat

y

as a constant

Apply the chain rule:

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

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

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

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

Simplify

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

= - \frac{2 y}{( x + y ) ( x - y )}

\int_{0}^{1} v^{2} cos (v^{3}) d v

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

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

\frac{d y}{d x} + y = e^{6 x}

\frac{d}{d x} (cos^{2} (sin (2 x)))

What is (\partial)/(\partial x)(ln((x+y)/(x-y))) ?
The answer to (\partial)/(\partial x)(ln((x+y)/(x-y))) is -(2y)/((x+y)(x-y))