Question

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

Answer 1

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

Explanation:

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)`