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Mathematics 16 Online
OpenStudy (anonymous):

derivative of. . . .

OpenStudy (anonymous):

\[f(x)=\cot \sqrt{x}\]

OpenStudy (anonymous):

you can avoid remembering too many things and use chain rule \[f(x)=\cot\sqrt{x}=(\tan\sqrt{x})^{-1}\\ f'(x)=(-1){1\over\tan^2\sqrt{x}}\times{d\over dx}(\tan\sqrt{x})\]

OpenStudy (anonymous):

but if you do, remember the differential of "cot", \[f'(x)=\csc^2\sqrt{x}\times{d\over dx}(\sqrt{x})\]

OpenStudy (anonymous):

I forgot a negative sign in the second one..

OpenStudy (anonymous):

thats ok. i know where thats suppose to be

OpenStudy (anonymous):

both ways, you'd get the same answer.

OpenStudy (anonymous):

so the answer i have is -csc^2 sqrt x /2 sqrt x

OpenStudy (anonymous):

yep thats the one

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