y=lncos(根号x)求y憋 设y=cos根号x+lnx,求dy
\u8bbey\uff1dln(cos^2x\uff0b\u6839\u53f7(1\uff0bcos^4x)\uff09\uff0c\u6c42dy=(-2cosxsinx+(-4cos³xsinx)/2\u221a(1+cos^4x))/(cos²x+\u221a(1+cos^4x))
=-sin2x/\u221a(1+cos^4x)
\u53c2\u8003
将三式相乘就是答案:y'=-tan(根号x)/(2根号x)
希望采纳!!!
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