# Is there an easy way to memorize the derivatives of inverse trigonometric functions?

Apr 7, 2015

The way is not to memorize.
The easiest way is to derive the formulae.
For e.g

$y = {\cos}^{-} 1 \left(x\right)$
then
$x = \cos y$
$\mathrm{dx} = - \sin y \mathrm{dy}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sin} \left(y\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\sqrt{1 - {x}^{2}}}$

Apr 10, 2015

There is some way to memorize, though you shouldn't have to. Knowing how to derive it is helpful for figuring them out again (maybe your internet goes out... I dunno). You can do it in "regular" and "co-" pairs.

While $\frac{d \left(a r c \tan x\right)}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}}$,

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

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

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

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

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

Now you only have to remember three of them. "co-" implies a reversed sign.