How do you differentiate #f(x)= 1/ (lnx)# using the quotient rule?

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Answer 1 · 2016-01-18T18:59:37

#f'(x)=-1/(xln^2(x))#

Given

#f(x)=(h(x))/g(x)#

#:.f'(x)=(h'(x)*g(x)-g'(x)*h(x))/[g(x)]^2#

In the question:

#h(x)=1=>h'(x)=0#
#g(x)=ln(x)=>g'(x)=1/x#

Substituting in the formula:

#f'(x)=(0*lnx-1/x*1)/[lnx]^2=-(1/x)/[lnx]^2=-1/(xln^2(x))#