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find the derivative of 1/tan(2x)
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look in your book for the derivative of cotangent
followed by chain rule :)
cotangent is -csc^2. so far it would be \[\frac{ 2\csc ^{2}x }{ (\tan2x)^{2} }\]
After that i get kind of confused on how to solve the rest.
hold the phone what is the derivative of cotangent?
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-csc^2u
actually all you have to remember is the derivative of tangent
since the derivative of tangent is secant squared, you get the derivative of cotangent is minus cosecant squared
in this case you have \(\cot(2x)\) and its derivative is therefore \[-2\csc^2(2x)\]
but when you use the chain rule it would turn positive because of the subtraction sign. u'v-uv'
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