Question

If $t(x)=ln(\frac{x}{x^2+1}),$ find $t^{\text{'}}(x).$

Select the correct answer below:

$t^{\text{'}}(x)=\frac{1}{x}-\frac{x^2+1}{2x}$

$t^{\text{'}}(x)=x-\frac{2x}{x^2+1}$

$t^{\text{'}}(x)=\frac{1}{x}+\frac{2x}{x^2+1}$

$t^{\text{'}}(x)=\frac{1}{x}-\frac{2x}{x^2+1}$

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