How do you differentiate $g(x) = 1/sqrtarctan(x^2-1)$ ?

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Answer 1 · 2018-06-10T04:20:15

$g'(x) = (-x)/(arctan^(3/2)(x^2-1)((x^2-1)^2+1)$

$g(x) = 1/(sqrt((arctan(x^2-1))$

$= [arctan(x^2-1)]^(-1/2)$

$g'(x) -1/2[arctan(x^2-1)]^(-3/2) * d/dx arctan(x^2-1)$

Apply standard derivative and chain rule

$g'(x) -1/2[arctan(x^2-1)]^(-3/2) * 1/(((x^2-1)^2+1)) * d/dx (x^2-1)$

$= -1/2[arctan(x^2-1)]^(-3/2) * 1/(((x^2-1)^2+1)) * 2x$

$= (-x)/[arctan^(3/2)(x^2-1)((x^2-1)^2+1)]$

How do you differentiate g(x) = 1/sqrtarctan(x^2-1) g(x) = 1/sqrtarctan(x^2-1) ?