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 )}

\frac{d}{d x} ((4 x + 1)^{2})

\frac{6}{s ^{2} + 5 s + 4}

\frac{d}{d x} (8 4 x^{2} + 7)

\int_{- 4}^{0} \frac{1}{( x + 2 ) ^{\frac{1}{3}}} d x

\frac{d}{d x} (\frac{( 6 e ^{x} )}{2 e ^{x} + 1})

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))