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Question #f0b82

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Aug 6, 2017

$d/dx(logx^2-cot2x) = 2/x+2csc^2 2x$

Explanation:

$d/dx(logx^2-cot2x) = d/dx(logx^2)-d/dx(cot2x)$

$d/dx(logx^2) = ((x^2)')/x^2 = (2x)/x^2=2/x$

$-d/dx(cot2x) = --2csc^2 2x= 2csc^2 2x$

$therefore d/dx(logx^2-cot2x) = 2/x+2csc^2 2x$

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