How do you differentiate $ln(1/(x^2+9))$ ?

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Answer 1 · 2016-06-14T06:57:26

$d/dxln(1/(x^2+9))=-(2x)/(x^2+9)$

As $f(x)=ln(1/(x^2+9))=ln1-ln(x^2+9)=0-ln(x^2+9)=-ln(x^2+9)$

Hence $(df)/(dx)=-1/(x^2+9)xx(2x)=-(2x)/(x^2+9)$

How do you differentiate ln(1/(x^2+9))ln(1/(x^2+9))?