Question

f(x)=ln[(x^2+1)/(x^2-1)] , find f'(x)

Answer 1

\[\frac{d}{dx} \ln(x^2+1) - \frac{d}{dx} \ln(x^2-1) \]
since ln(a/b) = lna-lnb

Answer 2

first one =\[\frac{2x}{x^2+1}\]
second one =\[\frac{2x}{x^2-1}\]

Answer 3

so\[f'(x) = \frac{2x}{x^2+1} - \frac{2x}{x^2-1}\]

Answer 4

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Answer 5

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Answer 6

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Answer 7

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Answer 8

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Answer 9

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