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Help me to solve the mentioned problem?

$\frac{d}{d x} (ln (e^{x} {(\frac{x - 1}{x + 1})}^{\frac{3}{2}}) = ?$ $d/(dx)(ln(e^x((x-1)/(x+1))^(3/2))=?$

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Answer 1 · 2017-12-05T11:52:33

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

Explanation:

First

$log (e^{x} {(\frac{x - 1}{x + 1})}^{\frac{3}{2}}) = x + \frac{3}{2} (log (x - 1) - log (x + 1))$ $log(e^x((x-1)/(x+1))^(3/2)) = x+3/2(log(x-1)-log(x+1))$ and then

$\frac{d}{d x} log (e^{x} {(\frac{x - 1}{x + 1})}^{\frac{3}{2}}) = 1 + \frac{3}{2} (\frac{1}{x - 1} - \frac{1}{x + 1}) = 1 + \frac{3}{x^{2} - 1} = \frac{x^{2} + 2}{x^{2} - 1}$ $d/(dx)log(e^x((x-1)/(x+1))^(3/2)) = 1+3/2(1/(x-1)-1/(x+1)) = 1+3/(x^2-1) = (x^2+2)/(x^2-1)$

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Help me to solve the mentioned problem?

$\frac{d}{d x} (ln (e^{x} {(\frac{x - 1}{x + 1})}^{\frac{3}{2}}) = ?$ $d/(dx)(ln(e^x((x-1)/(x+1))^(3/2))=?$

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Explanation:

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$\frac{d}{d x} (ln (e^{x} {(\frac{x - 1}{x + 1})}^{\frac{3}{2}}) = ?$ $d/(dx)(ln(e^x((x-1)/(x+1))^(3/2))=?$