How do you find the first and second derivative of $1/lnx$ ?

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Answer 1 · 2015-02-23T22:25:45

For the first derivative start by rewriting

$1/lnx=(lnx)^-1$

now take the derivative using power rule and chain rule

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

For the second derivative use the quotient rule. keep negative sign out in front so you do not lose track of it

$(d^2y)/dx^2=-[(x(lnx)^2(0)-((lnx)^2(1)+2(lnx)(1/x)x))/((x(lnx)^2)^2]]$

$(d^2y)/dx^2=-[(0-(lnx)^2-2ln(x))/(x^2(lnx)^4)]$

$(d^2y)/dx^2=((lnx)^2+2lnx)/(x^2(lnx)^4)$

$(d^2y)/dx^2=(lnx(lnx+2))/(x^2(lnx)^4)$

$(d^2y)/dx^2=(lnx+2)/(x^2(lnx)^3)$

How do you find the first and second derivative of 1/lnx1/lnx?