微積分上課筆記4

三角函數

$sin(sin^{-1}x)=x$
$\displaystyle \frac{d}{dx}sin(sin^{-1}x)=\frac{dx}{dx} = 1$
$\displaystyle =cos(sin^{-1}x) \frac{d}{dx}sin^{-1}x$
$\displaystyle (sin^{-1}x)’ = \frac{1}{cos(sin^{-1}x)}$
$\displaystyle =\frac{1}{\sqrt{cos^2(sin^{-1} x)}}$
$\displaystyle = \frac{1}{\sqrt{1 - sin^2(sin^{-1} x)}}$
$\displaystyle =\frac{1}{\sqrt{1 - x^2}}$

$cos$ 同理

$tan(tan^{-1}x)=x$
$\displaystyle \frac{d}{dx}tan(tan^{-1}x)=\frac{dx}{dx} = 1$
$\displaystyle =sec(tan^{-1}x) \frac{d}{dx}tan^{-1}x$
$\displaystyle =\frac{1}{1 + tan^2(tan^{-1} x)} = \frac{1}{1 + x^2}$

例題

Prove that $\displaystyle tan^{-1}x + cot^{-1} x = \frac{\pi}{2}$

solution

$\displaystyle (tan^{-1}x)’ + (cot^{-1})’ x = (\frac{\pi}{2})’ = 0$

考試範圍 8.2