Inverse Trigonometric Derivatives
Learn these:
$\displaystyle \frac{d}{dx} \sin^{-1}(x) =
\frac{1}{\sqrt{1-x^2}}$
$\displaystyle \frac{d}{dx} \tan^{-1}(x) =
\frac{1}{1+x^2}$
Ask your instructor if you also
need to learn this:
$\displaystyle \frac{d}{dx} \sec^{-1}(x) =
\frac{1}{x\sqrt{x^2-1}}$
We do not expect you to learn
these:
$\displaystyle \frac{d}{dx} \cos^{-1}(x) =
\frac{-1}{\sqrt{1-x^2}}$
$\displaystyle \frac{d}{dx} \csc^{-1}(x) =
\frac{-1}{x\sqrt{x^2-1}}$
$\displaystyle \frac{d}{dx} \cot^{-1}(x) =
\frac{-1}{1+x^2}$
|