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

The derivative of ln((1+x^2/(1-x^2))) is (4x)/((x^2+1)(-x^2+1))

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

The first derivative of ln((1+x^2/(1-x^2))) is (4x)/((x^2+1)(-x^2+1))

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

\frac{4 x}{( x ^{2} + 1 ) ( - x ^{2} + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{1 + x ^{2}}{1 - x ^{2}}) = \frac{4 x}{( 1 - x ^{2} ) ^{2}}

= \frac{1}{\frac{1 + x ^{2}}{1 - x ^{2}}} \cdot \frac{4 x}{( 1 - x ^{2} ) ^{2}}

Simplify

\frac{1}{\frac{1 + x ^{2}}{1 - x ^{2}}} \cdot \frac{4 x}{( 1 - x ^{2} ) ^{2}} : \frac{4 x}{( x ^{2} + 1 ) ( - x ^{2} + 1 )}

= \frac{4 x}{( x ^{2} + 1 ) ( - x ^{2} + 1 )}

\int \frac{1}{t ^{2} - 10 t + 29} d t

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

4 y^{^{''}} + 7 y = 0

\frac{\partial}{\partial x} (- 3 y)

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

What is the derivative of ln((1+x^2/(1-x^2))) ?
The derivative of ln((1+x^2/(1-x^2))) is (4x)/((x^2+1)(-x^2+1))
What is the first derivative of ln((1+x^2/(1-x^2))) ?
The first derivative of ln((1+x^2/(1-x^2))) is (4x)/((x^2+1)(-x^2+1))