How do you find the second derivative of $ln x/x^2$ ?

Question

5882 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T08:03:21

$(dy)/(dx)=(1-lnx^2)/x^3$

Here,

$y=lnx/x^2$

$"using"color(blue)" Quotient Rule"$

$(dy)/(dx)=(x^2d/(dx)(lnx)-lnxd/(dx)(x^2))/(x^2)^2$

$=(x^2*1/x-lnx(2x))/x^4$

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

$=(x(1-2lnx))/x^4$

$=(1-2lnx)/x^3$

$=(1-lnx^2)/x^3$

How do you find the second derivative of ln x/x^2ln x/x^2 ?