How do you find the derivative of $f (x) = ln (\frac{\sqrt[3]{x - 1}}{x + 1})$ $f(x)=ln(root3(x-1)/(x+1))$ ?

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Answer 1 · 2017-03-08T00:42:56

$f'(x)=frac{-2(x-2)}{3(x-1)(x+1)}$

The derivative of $lnu = (u')/u$
Using this fact, we can derive $f(x)$

$f(x)=ln(frac{root3(x-1)}{x+1})$

$f'(x)=frac{(frac{(x+1)(1/3)(x-1)^(-2/3)-(x-1)^(1/3)}{(x+1)^2})}{(frac{(x-1)^(1/3)}{x+1})}$

$f'(x)=(frac{cancel((x-1)^(1/3))(frac{x+1}{3(x-1)}-1)}{(x+1)^2})(frac{x+1}{cancel((x-1)^(1/3))})$

$f'(x)=frac{(frac{x+1-3x+3}{3(x-1)})}{x+1}$

$f'(x)=frac{-2x+4}{3(x-1)(x+1)}$

$f'(x)=frac{-2(x-2)}{3(x-1)(x+1)}$

How do you find the derivative of f(x)=ln(root3(x-1)/(x+1))f(x)=ln(3√x−1x+1)f(x)=ln(root3(x-1)/(x+1))?