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$\begingroup$

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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    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.2011-10-18

1 Answers 1

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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.

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    Oh after reading @J.M.'s [link](http://math.stackexchange.com/questions/30317) I think the answer for my prev comment is yes2011-10-19