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

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Answer 1 · 2018-06-11T20:17:10

$d/dx[ln(frac{x^2+1}{2x})] = frac{x^2-1}{x(x^2+1)}$

Use the chain rule. Also, note that $d/dx (lnx) = 1/x$

$d/dx[ln(frac{x^2+1}{2x})]$

$= frac{1}{(frac{x^2+1}{2x})} * frac{(2x)(2x)-(x^2+1)(2)}{(2x)^2}$

$= frac{2x}{x^2+1}*frac{4x^2 - 2x^2-2}{4x^2}$

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

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

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

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