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

Question

1833 views around the world

You can reuse this answer
Creative Commons License

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#