How do you find the derivative of Inverse trig function $y = a r c sec (\frac{1}{x})$ $y=arcsec(1/x)$ ?

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How do you find the derivative of Inverse trig function $y = a r c sec (\frac{1}{x})$ $y=arcsec(1/x)$ ?

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Answer 1 · 2015-07-09T03:35:48

$\frac{d}{d x} (a r c sec (\frac{1}{x})) = - \frac{1}{\sqrt{1 - x^{2}}}$ $d/dx(arcsec(1/x))=-1/sqrt{1-x^2}$ for $- 1 < x < 1$ $-1 < x < 1$ with $x \neq 0$ $x!=0$ .

Explanation:

First, let's find the derivative of $a r c sec (x)$ $arcsec(x)$ by applying the Chain Rule to the equation $sec (a r c sec (x)) = x$ $sec(arcsec(x))=x$ (and using the fact that $\frac{d}{d x} (sec (x)) = sec (x) tan (x)$ $d/dx(sec(x))=sec(x)tan(x)$ )

$\frac{d}{d x} (sec (a r c sec (x))) = \frac{d}{d x} (x) = 1 \Rightarrow sec (a r c sec (x)) tan (a r c sec (x)) \frac{d}{d x} (a r c sec (x)) = 1$ $d/dx(sec(arcsec(x)))=d/dx(x)=1\Rightarrow sec(arcsec(x))tan(arcsec(x))d/dx(arcsec(x))=1$

$\Rightarrow \frac{d}{d x} (a r c sec (x)) = \frac{1}{x \cdot \sqrt{x^{2} - 1}}$ $\Rightarrow d/dx(arcsec(x))=1/(x*sqrt{x^2-1})$ (draw a right triangle with one non-right angle labeled $a r c sec (x)$ $arcsec(x)$ and use the Pythagorean Theorem to help you do this last simplification).

Now the Chain Rule implies:

$\frac{d}{d x} (a r c sec (\frac{1}{x})) = \frac{d}{d x} (a r c sec (x^{- 1})) = \frac{- 1 x^{- 2}}{x^{- 1} \cdot \sqrt{{(x^{- 1})}^{2} - 1}}$ $d/dx(arcsec(1/x))=d/dx(arcsec(x^{-1}))=(-1x^{-2})/(x^{-1}*sqrt{(x^{-1})^2-1})$

This simplifies, after multiplying the top and bottom of the last fraction by $x^{2}$ $x^2$ , to:

$\frac{d}{d x} (a r c sec (\frac{1}{x})) = - \frac{1}{\sqrt{1 - x^{2}}}$ $d/dx(arcsec(1/x))=-1/sqrt{1-x^2}$

Evidently this implies that $a r c sec (\frac{1}{x}) = arccos (x) + C$ $arcsec(1/x)=arccos(x)+C$ for all allowed $x$ $x$ and some $C$ $C$ . In fact, it can be shown that $C = 0$ $C=0$ and $a r c sec (\frac{1}{x}) = arccos (x)$ $arcsec(1/x)=arccos(x)$ for all $- 1 \leq x \leq 1$ $-1\leq x\leq 1$ and $x \neq 0$ $x!=0$ .

Try graphing them both to check this!

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How do you find the derivative of Inverse trig function $y = a r c sec (\frac{1}{x})$ $y=arcsec(1/x)$ ?

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