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

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Answer 1 · 2017-01-04T08:48:05

$\frac{d y}{d x} = \frac{2}{x \sqrt{x^{4} - 1}}$ $(dy)/(dx)=2/(xsqrt(x^4-1))$

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

$\Rightarrow x^{2} = sec y$ $=>x^2=secy$

differentiate wrt $x$ $x$

$2 x = \frac{d y}{d x} sec y tan y$ $2x=(dy)/(dx)secytany$

$\frac{d y}{d x} = \frac{2 x}{sec y tan y}$ $(dy)/(dx)=(2x)/(secytany)$

now substitute back using:

$sec y = x^{2}$ $secy=x^2$

${sec}^{2} y = 1 + {tan}^{2} y \Rightarrow tan y = \sqrt{{sec}^{2} y - 1}$ $sec^2y=1+tan^2y=>tany=sqrt(sec^2y-1$

$tan y = \sqrt{x^{4} - 1}$ $tany=sqrt(x^4-1)$

$\frac{d y}{d x} = \frac{2 x}{x^{2} \sqrt{x^{4} - 1}} = \frac{2}{x \sqrt{x^{4} - 1}}$ $(dy)/(dx)=(2x)/(x^2sqrt(x^4-1))=2/(xsqrt(x^4-1))$

How do you differentiate y=sec^-1(x^2)y=sec−1(x2)y=sec^-1(x^2)?