What is d/dx(sec^-1(x^2)) for x > 1?

Question 43

MediumOriginal question

What is $\frac{d}{dx}\left(\sec^{-1}(x^{2})\right)$ for $x > 1$ ?

\frac{1}{x^{2}\sqrt{x^{4} - 1}}

Correct answerYour answer

\frac{2x}{\sqrt{x^{4} - 1}}

Correct answerYour answer

\frac{2}{x\sqrt{x^{4} - 1}}

Correct answerYour answer

\frac{2}{x\sqrt{x^{2} - 1}}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

$\frac{d}{dx}\sec^{-1} u = \frac{u'}{u\sqrt{u^{2} - 1}}$ for $u > 1$ , with $u = x^{2}$ , $u' = 2x$ .
$\frac{2x}{x^{2}\sqrt{x^{4} - 1}} = \frac{2}{x\sqrt{x^{4} - 1}}$ .

Worth noting

$\frac{1}{x^{2}\sqrt{x^{4} - 1}}$ puts $u = x^{2}$ into the formula but forgets to multiply by $u' = 2x$ .

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