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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

a^{2} x^{2} + b^{2}

y = (\frac{( x ^{2} + 8 )}{x ^{2} - 8})

3 t^{2} + 13 t + 13

e^{\frac{( x ^{5} )}{5}}

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