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Original question :

$$\frac{\tan(x)}{\log_e x}$$

I used the quotient rule and got the following fraction.

Fraction :

$$\frac{\frac {x \cdot \sec(x)^2 \cdot (\log_e x) - \tan(x)}{x}}{\log_e x}$$

apologies in advance for the fractions, i tried to put in $\log_e x$ underneath the denominator of the other fraction but for some reason it wasn't working.

I looked at the answer here:

https://www.symbolab.com/solver/step-by-step/%5Cfrac%7Bd%7D%7Bdx%7D%5Cleft(%5Cfrac%7B%5Ctan%20%5Cleft(x%5Cright)%7D%7B%5Clog%20_%7Be%7D%5Cleft(x%5Cright)%7D%5Cright)

My main issue is I nearly solved the whole thing, but towards the end they use this rule : $b/c/a = b/ca$. Before I solved this equation in the previous they used a this rule : $a/b/c = ac/b$. How do I identify which one is $a,b,c$. Because I used the rule from the previous equation and got my simplification wrong towards the end.

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    $\begingroup$ The notation "a/b/c" is ambiguous and should never be used! (a/b)/c= a/(bc) but a/(b/c)= ac/b. $\endgroup$ – user247327 Jan 13 '18 at 19:41
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Let $$f(x)=\frac{\tan x}{\ln x}$$ Using the Quotient Rule, we get $$f'(x)=\frac{\tan'x\cdot\ln x - \tan x \cdot \ln'x}{\ln^2x}=\frac{\frac{\ln x}{\cos^2 x}-\frac{\sin x}{x\cos x}}{\ln^2x}=\frac{x\ln x - \sin x \cos x}{x\cos^2x\ln^2x}$$ on multiplying numerator and denominator by $x\cos^2x$.

If you prefer to split it in fractions, $$f'(x)=\frac1{\ln x \cos^2x}-\frac{\tan x}{x\ln^2x}$$

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  • $\begingroup$ I looked at ur answer, how do you know which fraction rule you have to use from these : a/b/c=(a/b)/c=a/bc, a/b/c=a/(b/c)=ac/b? $\endgroup$ – jame_smith Jan 13 '18 at 19:53
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    $\begingroup$ It is the first one. $\endgroup$ – TheSimpliFire Jan 13 '18 at 19:54
  • $\begingroup$ but how would someone know which one to use? $\endgroup$ – jame_smith Jan 13 '18 at 20:00
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    $\begingroup$ Writing "a/b/c" plainly is very difficult to judge. But if you write $\frac{\frac ab}c$ then it would usually mean $\frac a{bc}$. $\endgroup$ – TheSimpliFire Jan 13 '18 at 20:02
  • $\begingroup$ oh ok, now it actually makes a lot more sense. thx a lot :) $\endgroup$ – jame_smith Jan 13 '18 at 20:03

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