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

1 Answer
Feb 10, 2016

Answer:

#f'(x) = (-2 ln x + 2)/(ln x)^2#

Explanation:

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#