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

Simplify

\frac{1}{\frac{1 - x ^{2}}{1 + x ^{2}}} (- \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{x}{( x ^{2} + 1 ) ^{\frac{9}{2}}} d x

- 4 (t - 1)

\int_{0}^{2} e^{5 x} d x

x \to 0 lim (\frac{x + 1}{x})

\int \frac{10 x ^{2} - 9 x + 4}{x ^{3} + 4 x} d x

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))