Differentiate $f(x) = lnabs(x/(1+x^2))$ ?

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Answer 1 · 2018-01-24T06:46:47

$f'(x) = 1/x - (2x)/(1+x^2)$

We must use our log laws:

$log_alpha(beta/gamma) -= log_alpha beta - log_alpha gamma$

$=> ln | x / (1+x^2) | = ln|x| - ln|1+x^2 |$

Now applying our knowledge of differentiating logs:

$d/(dx) ln(f(x)) = (f'(x))/f(x)$

By using the chain rule...

$d/(dx)( ln|x| - ln|1+x^2| ) = (d/(dx)(x))/x - (d/(dx)(1+x^2)) /( 1+x^2)$

$color(blue)(= 1/x - (2x)/(1+x^2)$

Differentiate f(x) = lnabs(x/(1+x^2))f(x) = lnabs(x/(1+x^2))?