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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Simplify

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

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

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} + 14 y = 0

\int \frac{12 x ^{3} + 6 x ^{2}}{3 x ^{4} + 2 x ^{3} - 5} d x

\frac{\partial}{\partial t} (1 + t)

f (x) = 4 + 7 x

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