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derivative of. . . .
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\[f(x)=\cot \sqrt{x}\]
you can avoid remembering too many things and use chain rule \[f(x)=\cot\sqrt{x}=(\tan\sqrt{x})^{-1}\\ f'(x)=(-1){1\over\tan^2\sqrt{x}}\times{d\over dx}(\tan\sqrt{x})\]
but if you do, remember the differential of "cot", \[f'(x)=\csc^2\sqrt{x}\times{d\over dx}(\sqrt{x})\]
I forgot a negative sign in the second one..
thats ok. i know where thats suppose to be
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both ways, you'd get the same answer.
so the answer i have is -csc^2 sqrt x /2 sqrt x
yep thats the one
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