How do you find the derivative of the function: $a r c sec (\frac{x}{2})$ $arcsec(x/2)$ ?

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How do you find the derivative of the function: $a r c sec (\frac{x}{2})$ $arcsec(x/2)$ ?

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Answer 1 · 2016-05-02T09:10:36

$\frac{d}{d x} arcsec (\frac{x}{2}) = \frac{2}{x \sqrt{x^{2} - 4}}$ $d/dx"arcsec"(x/2)=2/(xsqrt(x^2-4))$

Explanation:

Using implicit differentiation, we start by letting $y = arcsec (\frac{x}{2})$ $y = "arcsec"(x/2)$

$\Rightarrow sec (y) = \frac{x}{2}$ $=> sec(y) = x/2$

$\Rightarrow \frac{d}{d x} sec (y) = \frac{d}{d x} \frac{x}{2}$ $=> d/dxsec(y) = d/dxx/2$

$\Rightarrow sec (y) tan (y) \frac{d y}{d x} = \frac{1}{2}$ $=>sec(y)tan(y)dy/dx = 1/2$

$\Rightarrow \frac{d y}{d x} = \frac{1}{2 sec (y) tan (y)}$ $=> dy/dx = 1/(2sec(y)tan(y))$

We already know $sec (y) = \frac{x}{2}$ $sec(y) = x/2$ , and if we construct a right triangle with an angle $y$ $y$ such that $sec (y) = \frac{x}{2}$ $sec(y) = x/2$ we find that $tan (y) = \frac{\sqrt{x^{2} - 4}}{2}$ $tan(y) = sqrt(x^2-4)/2$ . Substituting these in, we have

$\frac{d}{d x} arcsec (\frac{x}{2}) = \frac{d y}{d x}$ $d/dx"arcsec"(x/2) = dy/dx$

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

$= \frac{2}{x \sqrt{x^{2} - 4}}$ $=2/(xsqrt(x^2-4))$

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How do you find the derivative of the function: $a r c sec (\frac{x}{2})$ $arcsec(x/2)$ ?

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