Differentiate y=sqrt(x+sqrt(x+(sqrt(x))) 求方程y=根号下x+根号x+根号x的导
Differentiate y=sqrt(x+sqrt(x+(sqrt(x))) 求方程y=根号下x+根号x+根号x的导数!
赞 李默深 幼苗
共回答了17个问题采纳率：88.2% 举报
dy={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} d[x+sqrt(x+(sqrt(x)))]
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {dx+1/[2sqrt(x)+(sqrt(x))]d[x+(sqrt(x))]}
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {dx+1/[2sqrt(x)+(sqrt(x))][dx+1/(2sqrt(x))dx]}
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {1+1/[2sqrt(x)+(sqrt(x))][1+1/(2sqrt(x))]}dx
即
Differentiate y={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {1+1/[2sqrt(x)+(sqrt(x))][1+1/(2sqrt(x))]}
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {dx+1/[2sqrt(x)+(sqrt(x))]d[x+(sqrt(x))]}
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {dx+1/[2sqrt(x)+(sqrt(x))][dx+1/(2sqrt(x))dx]}
={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {1+1/[2sqrt(x)+(sqrt(x))][1+1/(2sqrt(x))]}dx
即
Differentiate y={1/[2sqrt(x+sqrt(x+(sqrt(x)))]} {1+1/[2sqrt(x)+(sqrt(x))][1+1/(2sqrt(x))]}
1年前
2
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