How do you differentiate $y=sec^-1(x^2)$ ?

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Answer 1 · 2017-01-04T08:48:05

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

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

$=>x^2=secy$

differentiate wrt $x$

$2x=(dy)/(dx)secytany$

$(dy)/(dx)=(2x)/(secytany)$

now substitute back using:

$secy=x^2$

$sec^2y=1+tan^2y=>tany=sqrt(sec^2y-1$

$tany=sqrt(x^4-1)$

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

How do you differentiate y=sec^-1(x^2)y=sec^-1(x^2)?