What is the derivative of ln((x-1/(x+1))) ?

The derivative of ln((x-1/(x+1))) is 2/((x-1)(x+1))

What is the first derivative of ln((x-1/(x+1))) ?

The first derivative of ln((x-1/(x+1))) is 2/((x-1)(x+1))

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

\frac{2}{( x - 1 ) ( x + 1 )}

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

(x^{2} + 1) \frac{d y}{d x} + 2 x y - x = 0

x \to - 2 lim (\frac{x ^{2} + 8}{x + 2})

x^{^{''}} + 100 x = 0

\int x cos^{2} (x^{2}) d x

\frac{d y}{d t} - \frac{4 t}{1 + 4 t ^{2}} y = 4 t

What is the derivative of ln((x-1/(x+1))) ?
The derivative of ln((x-1/(x+1))) is 2/((x-1)(x+1))
What is the first derivative of ln((x-1/(x+1))) ?
The first derivative of ln((x-1/(x+1))) is 2/((x-1)(x+1))