2

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?

J. M. ain't a mathematician
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Jiew Meng
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    $\mathrm{arccos}$, aka $\cos^{-1}$, is the functional inverse of $\cos$, not the multiplicative reciprocal. – anon Oct 18 '11 at 08:37
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    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 08:37
  • @Ragib: You could write that as an answer; I don't think there's much more to say. – joriki Oct 18 '11 at 08:40
  • Agree with joriki in that Ragib (or anon) should write the comment as an answer. However, I would rather say that the author of the book is the confused one :-). At least I teach my calculus students to treat: $\cos^nx=(\cos x)^n$, and use $\arccos x$ for the inverse. As usual, the context should clear up this potential ambiguity. The lesson to be learned is that there is potential for ambiguity here. – Jyrki Lahtonen Oct 18 '11 at 08:49
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    @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 09:35
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    Very relevant. – J. M. ain't a mathematician Oct 18 '11 at 09:58
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    @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
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    The unfortunate notation $\cos^{-1}$ has been getting more common, not less! On no good evidence, I blame the calculator, on whose cramped keyboard $\arccos$ would not fit. Calculator manufacturers should use $\text{acos}$, which is common in computer languages. – André Nicolas Oct 18 '11 at 13:45

1 Answers1

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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.

mboratko
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anon
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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 08:48
  • Oh after reading @J.M.'s link I think the answer for my prev comment is yes – Jiew Meng Oct 19 '11 at 08:54