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I want to calculate the derivative of the following function $$ y = {(\cos x - 1)}^{\sec x - 1}$$

I searched and found a video on YouTube which started solving this by applying natural logarithm in both sides. $$ \ln (y) = \ln {(\cos x - 1)^{\sec x - 1}}$$ $$\Rightarrow \frac{y'}{y} = (\sec x - 1).[\ln(\cos x - 1)]$$ Here is the part I do not understand. Since we used natural logarithm we must have $$\cos x - 1> 0 \Rightarrow \cos x > 1$$

For logarithm to be well defined, however it is impossible for any x. So, I don't know whether the video from YouTube made a mistake or whether my reasoning has some flaw I can't find. Please help me understand the mistake and calculate the derivative properly! Thanks in advance!

K.K.McDonald
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Diego Nogueira Rocco
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    Indeed the exponential with negative base is not well defined. Isn't it $\cos(x-1)$? Can you give the link for the video? – user Jun 12 '23 at 20:01
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    Note that your statement after the $\implies$ sign is not correct because you have differentiated the left hand side but not the right hand side containing the contentious logarithm term – David Quinn Jun 12 '23 at 20:16
  • To have an easier algebra, you can substitute $u=\cos x$ but your problem of logarithm being non-definite will remain. – Ata Jun 12 '23 at 20:20
  • I guess you can still use the rule but the function $y$ is inherently complex – drC1Ron Jun 12 '23 at 20:21
  • The guy speaks Portuguese, so I guess you guys won't understand it (by the way, I've realized just now that I've written ''secx'', where it should be ''sinx''). Anyway, here's the link: https://youtu.be/x8D3RmWxhWM – Diego Nogueira Rocco Jun 12 '23 at 21:09
  • Why not just differentiate directly? – Adam Rubinson Jun 12 '23 at 21:50
  • Dear Diego, the video is wrong because the domain of the function, as written, is the set ${2\pi n : n \in \mathbb N}$, and you can't differentiate a function that isn't defined on an open set. The author's steps are correct, but that's assuming that the function he has is a differentiable function at all! There's a comment under the video now, which says that some kind of modulus has to be taken. The resulting calculation is unclear to me. Your reasoning has no problems, well done! – Sarvesh Ravichandran Iyer Jun 13 '23 at 12:10
  • I see now, thanks for the reply! – Diego Nogueira Rocco Jun 13 '23 at 23:20

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