How do you differentiate #f(x)=x/ln(sqrt(1/x))# using the chain rule?

Question

Evelyn Agard · Accepted Answer

#f'(x) = (-2 ln x + 2)/(ln x)^2# 1) Simplifying First of all, you can simplify using the following logarithmic laws: [1] #" "log_a(m/n) = log_a(m) - log_a(n)# [2] #" "log_a(m^r) = r * log_a(m)# and the power rule [3] #" "b^(1/2) = sqrt(b)# Thus, you have #ln(sqrt(1/x)) stackrel("[3] ")(=) ln [(1/x)^(1/2)] stackrel("[2] ")(=) 1/2 ln(1/x) stackrel("[1] ")(=) 1/2 (ln(1) - ln(x))# Now, #log_a(1) = 0# always holds for any basis #a > 0#, #a != 1#. Using this, you have, #ln(sqrt(1/x)) = 1/2 (ln(1) - ln(x)) = - 1/2 ln(x)# Now your function can be simplified to: #f(x) = x / ln(sqrt(1/x)) = (-2x) / ln(x)# 2) Differentiating It doesn't make sense to apply the chain rule to differentiate #f(x)# in this form. However, you can use the quotient rule: If #f(x) = (g(x)) / (h(x))#, then #f'(x) = (g'(x) h(x) - g(x) h'(x))/(h^2(x))# In your case, the derivatives of #g(x)# and #h(x)# are #g(x) = -2x " " =>" " g'(x) = -2# #h(x) = ln(x) " " =>" " h'(x) = 1/x# Thus, your derivative is: #f'(x) = (-2 * ln x + 2x * 1/x) / (ln x)^2 = (-2 ln x + 2)/(ln x)^2#