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y=√(x²+1/2x-1)
=>y'={1/[2√(x²+1/2x-1)]}(2x+1/2)
=(4x+1)/[4√(x²+1/2x-1)]
=>y'={1/[2√(x²+1/2x-1)]}(2x+1/2)
=(4x+1)/[4√(x²+1/2x-1)]
更多追问追答
追问
y=√(x²+1)/2x-1,求导数 有失误
追答
y=√(x²+1)/(2x-1) ?
y'={[(2x-1)x/√(x²+1)]-2√(x²+1)}/(2x-1)²
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