arcsin二分之x如何求导啊？ arcsin(x/2)的导数怎么导，有公式吗？要记住吗？谢谢...
y=arcsin(x/a)\u5982\u4f55\u6c42\u5bfc\uff0c\u6c42\u8be6\u7ec6\u89e3\u91ca\u590d\u5408\u51fd\u6570\u7684\u53cd\u51fd\u6570\u6c42\u5bfcy=arcsin(x/a) \u4e24\u8fb9\u53d6sin\uff1asiny=sin[arcsin(x/a)]=x/a\u4e24\u8fb9\u5bf9x\u6c42\u5bfccosy\u00b7y'=1/a\u2234y'=1/(acosy)=1/[a\u221a(1-sin²y)=1/a\u221a(1-x²)
\u6c42\u5bfc\u53cd\u51fd\u6570\u76f8\u5173\u95ee\u7b54
\u95ee\uff1a \u8bf7\u95ee\u4e00\u9053\u9ad8\u7b49\u6570\u5b66\u95ee\u9898
\u7b54\uff1ay=f(x)\u7684\u53cd\u51fd\u6570x=s(y)\u7684\u5bfc\u6570 s'(y)=1/(f'(x)) \u6ce8\u610fy\u548cx \u5982 \uff08arcsin(x))'=1/(sin(y))'=1/(cos(y))=1/sqrt(1-x^2) \u56e0\u4e3acos(y)=cos(arcsin(x))'=sqrt(1-x^2) (1/x)'=-1/x^2\u6ca1\u9519
\u7b54\uff1ay = f(x) \u7684\u53cd\u51fd\u6570\u662f x = g(y) y' = f'(x) = dy/dx x' = g'(y) = dx/dy \u6240\u4ee5\u6709 g'(x)f'(y) = 1 \u5bf9\u4f60\u8bf4\u7684 y = 1/x , f(x) = 1/x , g(y) = 1/y f'(x) = -1/x^2 g'(y) = -1/y^21...
\u95ee\uff1a \u53cd\u51fd\u6570\u6c42\u5bfc\u516c\u5f0f\u63a8\u5bfc \u539f\u51fd\u6570F(X)\u7684\u53cd\u51fd\u6570\u7684\u5012\u6570\u4e3a1/F'(X)\u662f\u600e\u4e48\u63a8\u5bfc\u51fa\u6765\u7684\uff1f
\u7b54\uff1a\u9996\u5148\u8981\u4fdd\u8bc1\u51fd\u6570y=f\uff08x\uff09\u5728\u5305\u542ba\u70b9\u7684\u5f00\u533a\u95f4I\u4e0a\u4e25\u683c\u5355\u8c03\u4e14\u8fde\u7eed\uff0c\u5982\u679c\u8fd9\u51fd\u6570\u5728a\u70b9\u53ef\u5bfc\u5e76\u4e14\u5bfc\u6570f'\uff08a\uff09\u22600\uff0c\u90a3\u4e48\u53cd\u51fd\u6570x=g\uff08y\uff09\u5728\u70b9b=f\uff08a\uff09\u53ef\u5bfc\uff0c\u4e14g'\uff08b\uff09=1/f'\uff08a\uff09=1/f'\uff08g\uff08b\uff09\uff09\u3002\u8bc1\u660e\uff1a\u5728\u6240\u7ed9\u6761\u4ef6\u4e0b\uff0c\u51fd\u6570x=g\uff08y\uff09\u4e5f\u4e25\u683c\u5355\u8c03\u4e14\u8fde\u7eed\u3002\u4e8e\u662f\uff0c\u5f53y\u2260b\uff0cy\u2192b\u65f6\uff0c\u6709g\uff08y\uff09\u2260g\uff08...
\u95ee\uff1a \u8bf7\u95ee\u7528\u53cd\u51fd\u6570\u6c42\u5bfc\u6cd5\u5219\u6c42SIN X\u7684\u53cd\u51fd\u6570\u7684\u5012\u6570\u600e\u4e48\u6c42\u5462\uff1f
\u7b54\uff1ay=sinx\u7684\u53cd\u51fd\u6570\u662fy=arcsinx,\u800c\u4f59\u5f26\u51fd\u6570\u6c42\u5bfc\u8fc7\u7a0b\u662f:y=cosx ==> y=sin(pi/2-x) ==> y'=[sin(pi/2-x)]'(-x)' ==> y'=-cos(pi/2-x) ==> y'=-sinx\u3002\u800cy=arcsinx, x=siny,(arcsinx)'=1/(...
\u95ee\uff1a \u53cd\u51fd\u6570\u6c42\u5bfc:
\u7b54\uff1a\u770b\u56fe```````````````
\u7b54\uff1a\u5229\u7528\u57fa\u672c\u516c\u5f0f\uff1a\uff081\uff09 (arcsinu)' = 1/\u221a(1-u²) \u3000\u3000\u3000\u3000\u3000\u3000\u3000\uff082\uff09 (\u221av)' = 1/(2\u221av) \u3000\u3000\u3000\u3000\u3000\u3000\u3000\uff083\uff09 (1-3x)' = -3 \u6839\u636e\u590d\u5408\u51fd\u6570\u7684\u6c42\u5bfc\u6cd5\u5219\uff0c\u5f97 y' = 1/\u221a(1-u²) * 1/(2\u221av) * (-3) \u5c06 u = \u221a(1...
arcsinx\u7684\u5bfc\u6570\u516c\u5f0f\u5c31\u662f
(arcsinx)'=1/\u221a(1-x²)
\u8fd9\u662f\u8981\u8bb0\u4f4f\u7684\u57fa\u672c\u516c\u5f0f
\u90a3\u4e48\u8fd9\u91cc\u5bf9arcsin(x/2)\u6c42\u5bfc
\u5f97\u5230(arcsinx/2)'=1/\u221a(1-x²/4) *(x/2)'
=1/\u221a(1-x²/4) *1/2
=1/\u221a(4-x²)
如图所示
参考资料：
arcsinx的求导网页链接
复合函数求导：网页链接
y= arcsin(x/2)
siny = x/2
cosy . y' = 1/2
y' = 1/(2cosy)
=1/ { 2√[ 1- (x/2)^2 ] }
=1/√(4- x)^2
供参考，请笑纳。待续
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