对y=sin[(cosx)^2]cos[(sinx)^2]求导 y=（sinx）^2求导
\u5df2\u77e5y=sin(cosx)^2*cos(sinx)^2,\u6c42y'y'=2*((-sin(x))*cos(cos(x)))*sin(cos(x))*cos(sin(x))^2+sin(cos(x))^2*(2*(-cos(x)*sin(sin(x)))*cos(sin(x)))
sinx^2\u7684\u5bfc\u6570\u662fsin2x\u3002
\u89e3\u7b54\u8fc7\u7a0b\u5982\u4e0b\uff1a
\u6269\u5c55\u8d44\u6599
\u79ef\u5316\u548c\u5dee\u516c\u5f0f\uff1a
sin\u03b1\u00b7cos\u03b2=(1/2)[sin(\u03b1+\u03b2)+sin(\u03b1-\u03b2)]
cos\u03b1\u00b7sin\u03b2=(1/2)[sin(\u03b1+\u03b2)-sin(\u03b1-\u03b2)]
cos\u03b1\u00b7cos\u03b2=(1/2)[cos(\u03b1+\u03b2)+cos(\u03b1-\u03b2)]
sin\u03b1\u00b7sin\u03b2=-(1/2)[cos(\u03b1+\u03b2)-cos(\u03b1-\u03b2)]
\u00b7\u548c\u5dee\u5316\u79ef\u516c\u5f0f\uff1a
sin\u03b1+sin\u03b2=2sin[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
sin\u03b1-sin\u03b2=2cos[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
cos\u03b1+cos\u03b2=2cos[(\u03b1+\u03b2)/2]cos[(\u03b1-\u03b2)/2]
cos\u03b1-cos\u03b2=-2sin[(\u03b1+\u03b2)/2]sin[(\u03b1-\u03b2)/2]
你好！利用乘积求导公式与复合函数求导公式计算，下图是计算过程。经济数学团队帮你解答，请及时采纳。谢谢！
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