正余弦函数及其反函数导数推导

全栈程序员-站长 • 2026年3月19日下午11:25 • 未分类 • 阅读 1

正余弦函数及其反函数导数推导

背景

三角函数求导 $sin x$

(s i n x)' = lim d x \to 0 s i n ( x + d x ) - s i n ( x ) d x

$(sin x)' = \lim_{dx\to0}\frac{sin(x+dx)-sin(x)}{dx}$

= lim d x \to 0 s i n ( x + d x 2 + d x 2 ) - s i n ( x + d x 2 - d x 2 ) d x

$= \lim_{dx\to0}\frac{sin(x+\frac{dx}{2}+\frac{dx}{2})-sin(x+\frac{dx}{2}-\frac{dx}{2})}{dx}$

= lim d x \to 0 2 c o s ( x + d x 2 ) s i n ( d x 2 ) d x

$= \lim_{dx\to0}\frac{2cos(x+\frac{dx}{2})sin(\frac{dx}{2})}{dx}$

= lim d x \to 0 c o s ( x + d x 2 ) s i n ( d x 2 ) d x 2

$= \lim_{dx\to0}\frac{cos(x+\frac{dx}{2})sin(\frac{dx}{2})}{ \frac{dx}{2}}$

= lim d x \to 0 c o s (x + d x 2) \cdot s i n ( d x 2 ) d x 2

$= \lim_{dx\to0}cos(x+\frac{dx}{2}) \cdot \frac{sin(\frac{dx}{2})}{ \frac{dx}{2}}$

= lim d x \to 0 c o s (x + d x 2) \cdot 1 = c o s x

$= \lim_{dx\to0}cos(x+\frac{dx}{2}) \cdot 1 = cos x$

三角函数求导 $cos x$

(c o s x)' = lim d x \to 0 c o s ( x + d x ) - c o s ( x ) d x

$(cos x)' = \lim_{dx\to0}\frac{cos(x+dx)-cos(x)}{dx}$

= lim d x \to 0 c o s ( x + d x 2 + d x 2 ) - c o s ( x + d x 2 - d x 2 ) d x

$= \lim_{dx\to0}\frac{cos(x+\frac{dx}{2}+\frac{dx}{2})-cos(x+\frac{dx}{2}-\frac{dx}{2})}{dx}$

= lim d x \to 0 - 2 s i n ( x + d x 2 ) s i n ( d x 2 ) d x

$= \lim_{dx\to0}\frac{-2sin(x+\frac{dx}{2})sin(\frac{dx}{2})}{dx}$

= lim d x \to 0 - s i n ( x + d x 2 ) s i n ( d x 2 ) d x 2 = - s i n x

$= \lim_{dx\to0}\frac{-sin(x+\frac{dx}{2})sin(\frac{dx}{2})}{\frac{dx}{2}} = -sin x$

三角函数求导 $arcsin x$

(s i n (a r c s i n x))' = c o s (a r c s i n x) \cdot (a r c s i n x)'

$(sin(arcsin x))' = cos(arcsin x) \cdot (arcsin x)'$

1 = c o s (a r c s i n x) \cdot (a r c s i n x)'

$1 = cos(arcsin x) \cdot (arcsin x)'$

(a r c s i n x)' = 1 c o s ( a r c s i n x ) = 1 1 - x 2 - - - - - \sqrt

$(arcsin x)' = \frac{1}{cos(arcsin x)} = \frac{1}{\sqrt{1-x^2}}$

三角函数求导 $arccos x$

(c o s (a r c c o s x))' = - s i n (a r c c o s x) \cdot (a r c c o s x)'

$(cos(arccos x))' = -sin(arccos x) \cdot (arccos x)'$

1 = - s i n (a r c c o s x) \cdot (a r c c o s x)'

$1 = -sin(arccos x) \cdot (arccos x)'$

(a r c c o s x)' = 1 - s i n ( a r c c o s x ) = - 1 1 - x 2 - - - - - \sqrt

$(arccos x)' = \frac{1}{-sin(arccos x)} = -\frac{1}{\sqrt{1-x^2}}$

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正余弦函数及其反函数导数推导