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

Feb 10, 2016

$f ' \left(x\right) = \frac{- 2 \ln x + 2}{\ln x} ^ 2$

Explanation:

1) Simplifying

First of all, you can simplify using the following logarithmic laws:

 $\text{ } {\log}_{a} \left(\frac{m}{n}\right) = {\log}_{a} \left(m\right) - {\log}_{a} \left(n\right)$

 $\text{ } {\log}_{a} \left({m}^{r}\right) = r \cdot {\log}_{a} \left(m\right)$

and the power rule

 $\text{ } {b}^{\frac{1}{2}} = \sqrt{b}$

Thus, you have

$\ln \left(\sqrt{\frac{1}{x}}\right) \stackrel{\text{ ")(=) ln [(1/x)^(1/2)] stackrel(" ")(=) 1/2 ln(1/x) stackrel(" }}{=} \frac{1}{2} \left(\ln \left(1\right) - \ln \left(x\right)\right)$

Now, ${\log}_{a} \left(1\right) = 0$ always holds for any basis $a > 0$, $a \ne 1$.

Using this, you have,

$\ln \left(\sqrt{\frac{1}{x}}\right) = \frac{1}{2} \left(\ln \left(1\right) - \ln \left(x\right)\right) = - \frac{1}{2} \ln \left(x\right)$

Now your function can be simplified to:

$f \left(x\right) = \frac{x}{\ln} \left(\sqrt{\frac{1}{x}}\right) = \frac{- 2 x}{\ln} \left(x\right)$

2) Differentiating

It doesn't make sense to apply the chain rule to differentiate $f \left(x\right)$ in this form. However, you can use the quotient rule:

If $f \left(x\right) = \frac{g \left(x\right)}{h \left(x\right)}$, then

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$

In your case, the derivatives of $g \left(x\right)$ and $h \left(x\right)$ are

$g \left(x\right) = - 2 x \text{ " =>" } g ' \left(x\right) = - 2$
$h \left(x\right) = \ln \left(x\right) \text{ " =>" } h ' \left(x\right) = \frac{1}{x}$

Thus, your derivative is:

$f ' \left(x\right) = \frac{- 2 \cdot \ln x + 2 x \cdot \frac{1}{x}}{\ln x} ^ 2 = \frac{- 2 \ln x + 2}{\ln x} ^ 2$