作业帮ln√x² y²=arctany÷x
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/08 11:15:49
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'=(lnlnx)'/lnlnx=(lnx)'/lnxlnlnx=1/xlnxlnlnx
1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x
你做的是对的y=ln(ln√x)y`=[lh(ln√x)]`=1/ln√x*(ln√x)`=(1/ln√x)*(1/√x)*(√x)`=(1/ln√x)*(1/√x)*(1/2)/√x=1/(2xln
y+arctany-x=0dy/dx+1/(1+y^2)dy/dx-1=0dy/dx(1+1/(1+y^2)=1dy/dx=(1+y^2)/(2+y^2)
y''=-(2+2y^2)/y
复合函数f(x)=lnxg(x)=ln[ln(x)]r(x)=ln{lnln(x)]}r'(x)=[1/lnln(x)]g'(x)=[1/lnln(x)][1/ln(x)]f'(x)=[1/lnln(
x+arctany=y两边对x求导有：1+y'/(1+y²）=y'整理得：y'=1+1/y²
设y=2arctan(y/x),求dy/dx,d²y/dx².设F(x,y)=y-2arctan(y/x)=0,则dy/dx=-(∂F/∂x)/(ͦ
答案在插图：
y=(ln(ln(x))'/ln(ln(x))=(ln(x))'/(ln(x)(ln(ln(x)))=1/(xln(x)ln(ln(x)))
见图再问：不好意思啊~题目看错了，题目如图啊~
1.y=arcsin(cosx)y'=[1/√（1－cos²x）](-sinx)=-sinx√（1－cos²x）/sin²x＝－|sinx|/sinx∴当sinx＞0时y
是(arctany)/x还是arctan(y/x)?如果是z=(arctany)/x,则∂z/∂x=-(arctany)/x²∂z/∂y=1/
x≤0时√x^2=-x所以y=0x>0时√x^2=x所以y=ln（2x+1）
利用查表或反函数求导法可求得(arctanu)'=1/(1+u^2)∴上述方程两边分别对x求导可得1+y'=y'/(1+y^2)=>(1+y^2)+(1+y^2)y'=y'=>(1+y^2)+y^2y
左右2边取正切,左边=(X+Y)/(1-XY)=右边.左边=arctan[(X+Y)/(1-XY)+Z]/[1-(X+Y)Z/(1-XY)]=arctanc(X+Y+Z-XYZ)/[1-XY-(X+Y
❶证明：tan(arctanX+arctanY)=(X+Y)/(1-XY)证明：tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tana
全微分吗?z=arctan(y/x)∂z/∂x=1/(1+y²/x²)*y=x²y/(x²+y²)∂z/ͦ