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

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

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

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

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derivative of

f (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}} (- \frac{2}{( x - 1 ) ^{2}})

Simplify

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

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

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

5 x^{2} + 4 x + 2

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (a x)

2 x^{3} + 3 x^{2} - 12 x + 5

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

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