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The correct way seems to be

$$\frac{d}{dx} \cos^{-1}{(\ln{x})} = \frac{ \frac{-1}{x} }{\sqrt{1-(\ln{x})^2}} = - \frac{1}{x \sqrt{1-(\ln{x})^2}}$$

But why not

$\frac{d}{dx} \cos^{-1}{(\ln{x})} = \frac{d}{dx} (\cos{(\ln{x})})^{-1} = -1 (\cos{(\ln{x})})^{-2}(- \frac{1}{x} \sin{(\ln{x})}) = \frac{\sin{(\ln{x})}}{x \cdot \cos^2{(\ln{x})}}$

Is it wrong? Maybe its another careless mistake again?

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$\mathrm{arccos}$, aka $\cos^{-1}$, is the functional inverse of $\cos$, not the multiplicative reciprocal. – anon Oct 18 '11 at 8:37
It's because of the confusing and unfortunate notation. The problem intends you to read $ \cos^{-1} (t) = \arccos t $, the *inverse* function of $\cos t$, but indeed the notation is often read by confused calculus students as $\cos^{-1} (t) = 1/(\cos t) $. So it is not really a careless mistake on your part, but it's something you will have to look out for, because it is a common notation. – Ragib Zaman Oct 18 '11 at 8:37
@joriki I just got back to the computer, and it seems anon's answer is sufficient. It seems like I need to become less hesitant in entering my input as an answer, it just feels a bit odd to do so when it's only a few lines that could slip into the comments. – Ragib Zaman Oct 18 '11 at 9:35
Very relevant. – J. M. Oct 18 '11 at 9:58
@Ragib: Yes -- I used to feel the same way, but the very high proportion of questions that remain formally unanswered and clutter the system because they were answered in the comments has convinced me otherwise. If you don't want to earn points for answers you consider trivial, you can always mark the answer as community wiki. – joriki Oct 18 '11 at 10:47
up vote 10 down vote accepted

The function $\cos^{-1}$ sometimes stands for the functional inverse of the cosine function (as it does here, apparently), so that it obeys the composition law $\cos(\cos^{-1}(x))=x$. It is not the multiplicative inverse of the function, or what we would write as $1/\cos x$. (Usually it is better to write $\arccos$ instead of $\cos^{-1}$ because it isn't so potentially ambiguous, just heads up.)

So you cannot use $\cos^{-1}\circ=(\cos\circ)^{-1}$ in your derivation.

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The composition law should read $\cos(\cos^{-1}(x))=x$. The one you wrote only hold for some $x$. – wckronholm Oct 18 '11 at 14:57
Hmm, but I suppose other expressions like $cos^2{x}$ I could differenciate it like "normal"? $\frac{d}{dx} \cos^2{x} = \frac{d}{dx} (\cos{x})^2 = 2 \cos{x} (- \sin{x})$ – Jiew Meng Oct 19 '11 at 8:48
Oh after reading @J.M.'s link I think the answer for my prev comment is yes – Jiew Meng Oct 19 '11 at 8:54

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