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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

\frac{\partial}{\partial x} (t x e^{\frac{x ^{2}}{2 t}})

x^{4} - 9 x^{2}, x^{2} - 9

2 csc (x)

\frac{\partial}{\partial x} (\frac{1}{2} ln (x^{2} + y^{2}))

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