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

##### 1 Answer

#### 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,

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

If

#f'(x) = (g'(x) h(x) - g(x) h'(x))/(h^2(x))#

In your case, the derivatives of

#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#