I am trying to show that $f(x) = \dfrac{x}{(\ln x)^2}$ is an increasing function for $x \ge 10$
Please let me know if any of these steps are wrong:
(1) Using the quotient rule with $g(x) = x$ and $h(x) = (\ln x)^2$:
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$
(2) Using the exponent rule for derivatives with $s(x) = \ln x$:
$$h'(x) = (s(x)^2)' = s(x)^2\left(s'(x)\dfrac{2}{s(x)}\right) = 2s(x)s'(x)$$
(3) $s'(x) = \dfrac{1}{x}$ so that:
$$h'(x) = \dfrac{2\ln(x)}{x}$$
(4) With $g'(x) = 1$, it follows that:
$$f'(x) = \dfrac{(\ln x)^2 - \frac{2x\ln(x)}{x}}{(\ln x)^4} = \dfrac{(\ln x)^2 - 2\ln(x)}{(\ln x)^4}$$
(5) It is increasing at $x=10$ since:
$$\dfrac{(\ln 10)^2 - 2\ln(10)}{(\ln(10))^4} > 0.24 > 0$$
Are these steps correct? Did I make any mistakes? Am I missing any details?
Easier approach: Introduce substitution $x=e^t$. Notice that $x(t)$ is monotoniously increasing. Now you have to prove that the following function is increasing:
$$f(t)=\frac{e^t}{t^2}$$
This is obvious from:
$$f'(t)=\frac{t^2e^t-2te^t}{t^4}=te^t\frac{t-2}{t^4}$$
The first derivative is positive for $t>2$ i.e. for $x>e^2\approx 7.39$.