Question #b5d1e

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Answer 1 · 2017-10-29T15:28:36

$2xcscx - 2x^2cot2xcsc2x$

We can approach this using the quotient rule;

$y= x^2csc2x$ = $x^2/(sin2x)$

hence quoteint rule; if $f(x)=u/v, then ,f'(x) = (vdu-udv)/v^2$

then $d/dx(x^2/(sin2x)) = (sin2x*2x - x^2*2cos2x)/sin^2(2x)$

as $d/dx( sinlamdax ) = lamdacos(lamdax)$

Hence simplifying to give;

$dy/dx = 2xcsc2x - 2x^2cot2xcsc2x$