cos^2(1/x)的导数怎么求 y=2^cos 1/x 导数求
cos^2\uff081/x\uff09\u7684\u5bfc\u6570\u600e\u4e48\u6c42\u7c7b\u4f3c\u7684\u5462\u5c061/x\u770b\u505a\u662fv\uff0c\u5c06cos(1/x)\u770b\u505a\u662fu\uff1a
\u539f\u5f0f = cos²v = u²
\u6839\u636e\u5206\u6b65\u6c42\u5bfc\uff1a
\uff08u²\uff09 = 2u* u\u2032
= 2cosv * (cosv)\u2032
= 2cosv * (-sinv) * v\u2032
= -sin(2v) * v\u2032
= -sin(2/x) * (1/x)\u2032
= -sin(2/x) * (-1/x²)
= sin(2/x) / x²
y=2^cos1/x
y'=2^cos1/x*ln2*(cos1/x)'=2^cos1/x*ln2*(-sin1/x)*(1/x)'=ln2*2^cos1/x*sin1/x*1/x^2
自外到内，链式法则。
I'=-2cos(1/x)sin(1/x)(-1/x^2)
=2cos(1/x)sin(1/x)/x^2
绛旓細灏1/x鐪嬪仛鏄痸锛屽皢cos(1/x)鐪嬪仛鏄痷锛氬師寮 = cos²v = u²鏍规嵁鍒嗘姹傚锛氾紙u²锛 = 2u* u鈥= 2cosv * (cosv)鈥= 2cosv * (-sinv) * v鈥= -sin(2v) * v鈥= -sin(2/x) * (1/x)鈥= -sin(2/x) * (-1/x²)= sin(2/x) / x²
绛旓細Y' = 2cos(1-x)路(-1)路sin(1-x)路(-1)= 2cos(1-x)sin(1-x)娉ㄦ剰cosx姹傚寰楋細 鈥 sinx
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