What is the second derivative of # y=x/lnx#?

Question

Charles Anctil · Accepted Answer

#(d^2y)/dx^2 = (2-ln(x))/(x(ln(x))^3)#

To find the first derivative we must use the quotient rule:

#(d(g/h))/dx = (g'(h)-g(h'))/h^2#

where #g = x and h = ln(x)#, then:

#g'=1 and h' = 1/x#

Substituting into the quotient rule:

#(d(x/ln(x)))/dx = (1ln(x)-x(1/x))/(ln(x))^2#

#(d(x/ln(x)))/dx = (ln(x)-1)/(ln(x))^2#

The second derivative, also, requires the quotient rule:

#(d(g/h))/dx = (g'(h)-g(h'))/h^2#

where #g = ln(x)-1 and h = (ln(x))^2#, then:

#g'=1/x and h' = (2ln(x))/x#

Substituting into the quotient rule:

#(d((ln(x)-1)/(ln(x))^2))/dx = (1/x(ln(x))^2-(ln(x)-1)((2ln(x))/x))/(ln(x))^4#

#(d((ln(x)-1)/(ln(x))^2))/dx = (1/x(ln(x))^2-2/x(ln(x))^2+2/xln(x))/(ln(x))^4#

#(d((ln(x)-1)/(ln(x))^2))/dx = (2-ln(x))/(x(ln(x))^3)#

This is the second derivative of the original function:

#(d^2y)/dx^2 = (2-ln(x))/(x(ln(x))^3)#

Christopher Agresta · Answer

#d^2y/dx^2=(2-lnx)/(x(lnx)^3)# differentiate using the #color(blue)"quotient rule"# #"Given " y=(g(x))/(h(x))" then"# #• dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2larr" quotient rule"# #"here " g(x)=xrArrg'(x)=1# #"and " h(x)=lnxrArrh'(x)=1/x# #rArrdy/dx=(lnx-x . 1/x)/(lnx)^2# #color(white)(rArrdy/dx)=(lnx-1)/(lnx)^2# differentiate again using the #color(blue)"quotient rule"# #g(x)=lnx-1rArrg'(x)=1/x# #h(x)=(lnx)^2rArrh'(x)=2lnx . 1/x=2/xlnx# #rArr(d^2y)/dx^2=(1/x(lnx)^2-2/xlnx(lnx-1))/(lnx)^4# #color(white)(rArrd^2y/dx^2)=(1/x((lnx)^2-2lnx(lnx-1)))/(lnx)^4# #color(white)(rArrd^y/dx^2)=(1/x((lnx)^2-2(lnx)^2+2lnx))/(lnx)^4# #color(white)(rArrd^2y/dx^2)=(lnx(2-lnx))/(x(lnx)^4# #color(white)(rArrd^2y/dx^2)=(2-lnx)/(x(lnx)^3)#