求导:y=ln[x+(a²+x²)½]
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y'=ln[x+(a²+x²)½]'
=1/[x+(a²+x²)½]×[x+(a²+x²)½]'
=[x+(a²+x²)½]'/[x+(a²+x²)½]
=[1+((a²+x²)½)']/[x+(a²+x²)½]
=[1+(1/2(a²+x²)^(½-1)(a^2+x^2)']/[x+(a²+x²)½]
=[1+1/2(a²+x²)^(-½)×2x]/[x+(a²+x²)½]
=[1+(a²+x²)^(-½)x]/[x+(a²+x²)½]
=1/[x+(a²+x²)½]×[x+(a²+x²)½]'
=[x+(a²+x²)½]'/[x+(a²+x²)½]
=[1+((a²+x²)½)']/[x+(a²+x²)½]
=[1+(1/2(a²+x²)^(½-1)(a^2+x^2)']/[x+(a²+x²)½]
=[1+1/2(a²+x²)^(-½)×2x]/[x+(a²+x²)½]
=[1+(a²+x²)^(-½)x]/[x+(a²+x²)½]
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与标准答案不符啊😂答案是y'=(a²+x²)^-½😂求大神再算一下🙏🙏🙏
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