How do you find the derivitive of Inverse trig function $y=cot(1-2(x)^2)$ ?

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Answer 1 · 2015-06-17T16:43:57

$(4x)/(1+4x^4)$

Given function is y= $cot^-1 2x^2$

Cot y= $2x^2$ . Now differentiate w.r.t x, to have $sec^2y dy/dx=4x$

$dy/dx= (4x)/(1+cot^2 y)= (4x)/(1+4x^4)$

How do you find the derivitive of Inverse trig function y=cot(1-2(x)^2)y=cot(1-2(x)^2)?