I have completed calculus 1 but every so often I still get a litte confused on some basic derivatives. I will give an example, I am not sure if my answer is correct but I will show my work.

$ d/dx$ for $f(x)=\frac{\ln x}{\sqrt{x}}$

I used the quotient rule, so I got $\frac{\sqrt{x}(1/x)-(\ln x)((1/2)x^{-1/2})}{x}$

I think I am having trouble simplifying. Because I am trying to use this limit so that I can use the alternating series test so I need to show that the function is decreasing.

Here is what I attempted for the simplification

I wrote $f'(x)=((\sqrt x/x)-\ln x)/(2x \sqrt x )=\sqrt x-\ln x/(2x^{5/2})$

But I am not sure if this is correct or if it even allows me to conclude decreasing. Any help suggestions or anything is very appreciated. Thank you all

  • $\begingroup$ You don't have to use \$\$ around every part of an equation. You can just put \$\$ around an entire equation. Functions like $\ln$ have a general command in mathmode by \ln. $\endgroup$ Feb 23 '15 at 2:43

You have

$$f'(x)=\frac{\sqrt{x}\cdot\frac1x-(\ln x)\frac12x^{-1/2}}x\;,$$

which is correct. Multiplying numerator and denominator by $x^{1/2}$ will clean up both terms in the numerator quite a bit: it will turn the $\sqrt{x}$ in the first term into $x$, which will cancel with the $\frac1x$, and it will get rid of the $x^{-1/2}$ in the second term. We now have

$$f'(x)=\frac{1-\frac12\ln x}{x^{3/2}}\;.$$

Multiplying top and bottom by $2$ looks like a good idea now:

$$f'(x)=\frac{2-\ln x}{2x^{3/2}}\;.$$

Now it should be much easier to decide where $f'(x)$ is negative.


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