How do you find the derivative $g (x) = a r c sec (2 x)$ $g(x)=arcsec(2x)$ ?

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How do you find the derivative $g (x) = a r c sec (2 x)$ $g(x)=arcsec(2x)$ ?

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Answer 1 · 2017-12-29T12:17:27

$\frac{d}{d x} ({sec}^{- 1} (2 x)) = \frac{1}{| x | \sqrt{4 x^{2} - 1}}$ $d/dx(sec^(-1)(2x)) = 1/(|x|sqrt(4x^2-1)$

Explanation:

We know that $\frac{d}{d x} ({sec}^{- 1} (x)) = \frac{1}{| x | \sqrt{x^{2} - 1}}$ $d/dx(sec^-1(x)) = 1/(|x|sqrt(x^2-1))$
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In this case, $\frac{d}{d x} ({sec}^{- 1} (2 x)) = \frac{1}{| 2 x | \sqrt{{(2 x)}^{2}} - 1} \cdot 2$ $d/dx(sec^(-1)(2x)) = 1/(|2x|sqrt((2x)^2)-1)*2$ .

We can rewrite this a bit, since $| 2 x | = 2 \cdot | x |$ $|2x| = 2*|x|$ , the 2 from here will cancel with the 2 we got from the chain rule. Also, ${(2 x)}^{2} = 4 x^{2}$ $(2x)^2=4x^2$ . So we end up with:

$\frac{d}{d x} ({sec}^{- 1} (2 x)) = \frac{1}{| x | \sqrt{4 x^{2} - 1}}$ $d/dx(sec^(-1)(2x)) = 1/(|x|sqrt(4x^2-1)$ .

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How do you find the derivative $g (x) = a r c sec (2 x)$ $g(x)=arcsec(2x)$ ?

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How do you find the derivative $g (x) = a r c sec (2 x)$ $g(x)=arcsec(2x)$ ?