What is the derivative of #f(x) = 1/(xlnx)#?

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Answer 1 · 2018-03-22T14:18:23

#f^'(x)=-(1+lnx)/(xlnx)^2#

We know that,
#color(red)((1) d/(dx)(X^n)=nX^(n-1)#
#color(red)((2) d/(dx)(u*v)=u*(dv)/(dx)+v*(du)/(dx)#

Here,
#f(x)=1/(xlnx)=(xlnx)^-1#

Using chain rule and (1)

#f^'(x)=-1*(xlnx)^(-1-1)d/(dx)(xlnx)#

#=-(xlnx)^-2[xd/(dx)(lnx)+lnxd/(dx)(x)],..to#Apply(2)

#=-1/(xlnx)^2[x*1/x+lnx(1)]#

#=-1/(xlnx)^2[1+lnx]#

#=-(1+lnx)/(xlnx)^2#