What is the derivative of this function $y=sec^-1(x^2)$ ?

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Answer 1 · 2017-10-24T11:55:31

$dy/dx= 2/(xsqrt(x^4-1)$

$y = sec^-1 (x^2)$

$y = arccos(1/x^2)$

Apply standard differential and chain rule

$dy/dx= (-1/sqrt(1-(1/x^2)^2)) * d/dx (1/x^2)$

$dy/dx = -1/sqrt(1-1/x^4) * -2/x^3$

$= 1/sqrt(x^4-1) * sqrt(x^4) * 2/x^3$

$= 1/sqrt(x^4-1) * x^2 * 2/x^3$

$= 2/(xsqrt(x^4-1)$

What is the derivative of this function y=sec^-1(x^2)y=sec^-1(x^2)?