What is the derivative of ln((x+y/(x-y))) ?

The derivative of ln((x+y/(x-y))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

What is the first derivative of ln((x+y/(x-y))) ?

The first derivative of ln((x+y/(x-y))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial s} (sin (s^{2} - t^{2}))

\frac{d}{d x} (ln (2 x^{4} e^{3 x} (x^{2} + x + 1)^{5}))

\int \frac{3 - x}{3 - x ^{2}} d x

\int_{- 2}^{5} - 2 d x

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

What is the derivative of ln((x+y/(x-y))) ?
The derivative of ln((x+y/(x-y))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}
What is the first derivative of ln((x+y/(x-y))) ?
The first derivative of ln((x+y/(x-y))) is (2x\frac{dy)/(dx)-2y}{(x+y)(x-y)}