How do you find the first and second derivative of #(lnx)^(x)#?
1 Answer
Let:
#y=(lnx)^x#
The easiest way to differentiate this is to take the natural logarithm of both sides.
#ln(y)=ln((lnx)^x)#
Simplify using the logarithm rule:
#ln(y)=xln(lnx)#
Differentiate both sides. The left-hand side will need chain rule, similar to implicit differentiation, and the right-hand side will need chain rule and product rule.
#1/y*dy/dx=d/dx(x)*ln(lnx)+x*d/dxln(lnx)#
#1/y*dy/dx=ln(lnx)+x*1/lnx*d/dxlnx#
#1/y*dy/dx=ln(lnx)+x*1/lnx*1/x#
#1/y*dy/dx=ln(lnx)+1/lnx#
#1/y*dy/dx=(lnx*ln(lnx)+1)/lnx#
#dy/dx=(y(lnx*ln(lnx)+1))/lnx#
#dy/dx=((lnx)^x(lnx*ln(lnx)+1))/lnx#
#dy/dx=(lnx)^(x-1)(lnx*ln(lnx)+1)#
The steps to find the second derivative are far too lengthy, and the work is rather pointless, but if you wish try to find it, the second derivative is:
#(d^2y)/(dx^2)=((lnx)^(x-2)(lnx(xln(lnx)(lnx*ln(lnx)+2)+1)+x-1))/x#