# How do you find the derivative of Inverse trig function y=arcsec(1/x) ?

Jul 9, 2015

$\frac{d}{\mathrm{dx}} \left(a r c \sec \left(\frac{1}{x}\right)\right) = - \frac{1}{\sqrt{1 - {x}^{2}}}$ for $- 1 < x < 1$ with $x \ne 0$.

#### Explanation:

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

$\frac{d}{\mathrm{dx}} \left(\sec \left(a r c \sec \left(x\right)\right)\right) = \frac{d}{\mathrm{dx}} \left(x\right) = 1 \setminus R i g h t a r r o w \sec \left(a r c \sec \left(x\right)\right) \tan \left(a r c \sec \left(x\right)\right) \frac{d}{\mathrm{dx}} \left(a r c \sec \left(x\right)\right) = 1$

$\setminus R i g h t a r r o w \frac{d}{\mathrm{dx}} \left(a r c \sec \left(x\right)\right) = \frac{1}{x \cdot \sqrt{{x}^{2} - 1}}$ (draw a right triangle with one non-right angle labeled $a r c \sec \left(x\right)$ and use the Pythagorean Theorem to help you do this last simplification).

Now the Chain Rule implies:

$\frac{d}{\mathrm{dx}} \left(a r c \sec \left(\frac{1}{x}\right)\right) = \frac{d}{\mathrm{dx}} \left(a r c \sec \left({x}^{- 1}\right)\right) = \frac{- 1 {x}^{- 2}}{{x}^{- 1} \cdot \sqrt{{\left({x}^{- 1}\right)}^{2} - 1}}$

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

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

Evidently this implies that $a r c \sec \left(\frac{1}{x}\right) = \arccos \left(x\right) + C$ for all allowed $x$ and some $C$. In fact, it can be shown that $C = 0$ and $a r c \sec \left(\frac{1}{x}\right) = \arccos \left(x\right)$ for all $- 1 \setminus \le q x \setminus \le q 1$ and $x \ne 0$.

Try graphing them both to check this!