How do you find the first and second derivative of $ln sqrt (x^2-4)$ ?

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Answer 1 · 2017-04-25T12:23:02

$(df)/(dx)=x/(x^2-4)$ and $(d^2f)/(dx^2)=-(x^2+4)/(x^2-4)^2$

We can use chain rule here, for which first differentiate w.r.t. $sqrt(x^2-4)$ and ten with respect to $(x^2-4)$ and finally w.r.t. $x$ .

As such, as $f(x)=ln(sqrt(x^2-4))$

$(df)/(dx)=1/sqrt(x^2-4)xx1/(2sqrt(x^2-4))xx2x$

= $x/(x^2-4)$

It is apparent that for second derivative, we can use quotient formula and hence

$(d^2f)/(dx^2)=((x^2-4)xx1-x xx 2x)/(x^2-4)^2$

= $(-x^2-4)/(x^2-4)^2$

= $-(x^2+4)/(x^2-4)^2$

How do you find the first and second derivative of ln sqrt (x^2-4)ln sqrt (x^2-4)?