已知ln根号x^2 y^2=arctany x-搜答案-智慧作业帮

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)]

再答：��Ϻ��

y=√(1+ln^2*x)y'=[1/2√(1+ln^2x)]*(2lnx)*1/x则lnxy'=----------------------x√(1+ln^2x)

两边相加都是0,没啥意义啊,我有一种方法

求导y=ln ln ln(x^2+1)

=[1+x/(x^2+1)^(1/2)]/[x+(1+x^2)^(1/2)]

（根号y/根号x-根号y)-(根号y/根号x+根号y)=｛根号y（根号x+根号y）｝/（x-y）-｛根号y（根号x-根号y）｝/(x-y)=(y+y)/(x-y)因为x=2y所以原式=2y/y=2

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'=1/(x+√(1+x²))*(x+√(1+x²)'(x+√(1+x²)'=1+1/[2√(1+x²)]*(1+x²)'=1+2x/[2√(1+x

1)这两个函数对所有实数有定义；2）ln[-x+根号下(x^2+1)]=ln[1/(x+根号下(x^2+1))]=-ln[x+根号下(x^2+1)]

因为所以

1,y=ln(1-x)y'=1/(1-x)*(1-x)'=1/(1-x)*(-1)=1/(x-1);2,y=ln[1/√(1-x)]=-ln√(1-x)y'=-1/√(1-x)*[√(1-x)]'=-

直接写重要步骤：两端对x求导,化简,得y-y'x=2x+2y-y'y'=(y-2x)/(x+2y)两端再对x求导,化简,并将上一步结果代入,得y''=-10(x^2+y^2)/(x+2y)^3

y'=(1+x/√(1+x^2))/(x+√(1+x^2))=1/√(1+x^2)y''=-x/(1+x^2)^(3/2)

原式=√y/(√2y-√y)-√y/(√2y+√y)=√y/[√y(√2-1)]-√y/[√y(√2+1)]=1/(√2-1)-1/(√2+1)=(√2+1)/(√2+1)(√2-1)-(√2-1)/

y-x^2>01-y-x>=0所以x^2

y'=(1+4x^3)/(2x+2x^4)

y=根号ln(2+3x^2)

题全吗?不就是一个函数式吗：y=√(2+3x^2)再问：是全题y=√ln(2+3x^2)求函数的导数再答：那不就行了，y=√(2+3x^2)=(2+3x^2)^1/2所以y'=1/2(2+3x^2)^