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The function:

$$f(x) = \frac{1}{x(\ln x)^2}$$

may be rewritten as:

$$f(x) = (x^{-1})(\ln x)^{-2}$$

We begin by using the product rule for differentiaton:

$$[f(x)g(x)]' = (-x^{-2})(\ln x)^2 + (x^{-1})(-2)(\ln x)^3)(x^{-1})$$

and is simplified to:

$$f'(x) = {-\frac{1}{(x\ln x)^2} + \frac{2}{x^2(\ln x)^3}}$$

Is this a correct derivation of $f(x) = \frac{1}{x(\ln x)^2}$?

How to I determine whether the function $f(x)$ is continous, increasing/decreaseing and on what interval?

Answer: $f(x)$ is continuous, positive, decreasing on $[2,\infty]$.

Also:

Why do I not need to calculuate a derivative in this situation?

rae306
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cryptomath
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  • Please look at my edits. I changed things like $x ln x$ to $x\ln x$, coded as x\ln x. That is standard usage. The backslash on \ln has at least two effects: (1) it de-italicizes "$\ln$" and (2) it provides proper spacing before and after "$\ln$" in expressions like $x\ln x$. ${}\qquad{}$ – Michael Hardy Sep 14 '14 at 16:53

1 Answers1

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The domain of your function is $D=\{x\in\mathbb{R}\ |\ x>0\}$. The function $f(x)=\ln(x)^2$ is continuous, $g(x)=1/x$ is continuous as well and so is the composition $g(f(x))=\ln(x)^{-2}$. Making the product of two continuous functions is also continuous, so $g(x)g(f(x))=\frac{1}{x\ln(x)^2}$ is continuous.

Now, in general, if $f(x)>0$ and $f'(x)>0$, you have $f^2(x)>0$ and $\frac{d}{dx}f^2(x)=2f(x)f'(x)>0$.

For $x>1$, the function $f(x)=\ln(x)$ is positive and increasing, so also the function $\ln(x)^2$ has the same properties. Product of positive and increasing functions is positive an increasing; the function $x$ is positive and increasing for $x>1$, so $x\ln(x)^2$ is positive and increasing. The reciprocal of an increasing function is decreasing, so there you are.

marco trevi
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