函数y=根号9-x^2 ln{x-1}
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函数y=ln(x+根号(1+x^2))的奇偶性和单调性无关.y=f(x)=ln(x+√(1+x^2))=ln(x+√(1+x^2))(x-√(1+x^2))/(x-√(1+x^2))=ln1/ln(-
两边对x求导得1/[1+(y/x)^2]*(y/x)'=1/[ln(x^2+y^2)]*[ln(x^2+y^2)]'1/[1+(y/x)^2]*(y'x-y)/x^2=1/[2ln(x^2+y^2)]
y=ln[2+根号(x^2+4)]/xy=ln[2+根号(x^2+4)]-lnx所以y'=1/[2+根号(x^2+4)]*[2+根号(x^2+4)]'-1/x=[x/√(x^2+4)]/[2+√(x^
先确定定义域,R,关于原点对称f(-x)=㏑(-x+√(1+(-x)²))=㏑(√(1+x²)-x)=㏑(1/(√(1+x²)+x))=-㏑(√(1+x²)+x
因为y=√(9-x^2)/ln(x+2)所以:9-x^2≥0………………(1)ln(x+2)≠0……………(2)x+2>0…………………(3)由(1),有:x^2≤9,得:-3≤x≤3由(2),有:x
两边相加都是0,没啥意义啊,我有一种方法
y'=1/(x+根号下x^2+1)*(x+根号下x^2+1)'=1/(x+根号下x^2+1)*(1+x/根号下x^2+1)=1/(x+根号下x^2+1)*(根号下x^2+1+x)/根号下x^2+1=1
=[1+x/(x^2+1)^(1/2)]/[x+(1+x^2)^(1/2)]
y'=1/[x-√(x^2-1)]×[1-x/√(x^2-1)]=1/[x-√(x^2-1)]×[(√(x^2-1)-x)/√(x^2-1)]=-1/√(x^2-1)
定义域是(-∞,+∞)y'=1/√(1+x^2)>0,所以函数在定义域内是增函数
x-1>0x>1x^2-4x+3>0(x-3)(x-1)>0x>3orx3
symsx>>y=log(x+sqrt(1+x^2));>>simple(diff(y)ans=1/(1+x^2)^(1/2)>>y=log(2*x+sqrt(1+x^2));>>simple(dif
y=ln[x+√(1+x²)]∴y'=[x+√(1+x²)]'/[x+√(1+x²)]=[1+x/√(1+x²)]/[x+√(1+x²)]=[x+√(
y=1/2[ln(1+x^2)-ln(1-x^2)]y'=1/2[2x/(1+x^2)-(-2x)/(1-x^2)]=x/(1+x^2)+x/(1-x^2)=2x/(1-x^4)
y'=1/(x+√(1+x²))*(x+√(1+x²)'(x+√(1+x²)'=1+1/[2√(1+x²)]*(1+x²)'=1+2x/[2√(1+x
y=根号下1+ln(x^2)+e^(2x)y′=1/2(1+ln(x^2)+e^(2x))ˆ(-1/2)(2/x+2e^(2x))=(2/x+2e^(2x))/2√(1+ln(x^2)+e^
y'=arctanx加x/(1加x^2)-x/(1加x^2)=arctanx再问:有详细步骤吗?
y=u^(1/2)u=lnVV=lnpp=x^(1/2)
由题意可得:x^2-2x02x-1不等于1联立解得1/2
y=ln(x+根号x^2+1)e(y次方)=x+根号(x²+1)e(y次方)-x=根号(x²+1)两边平方,有:e(2y次方)-2xe(y次方)+x²=x²+1