arctany除以x=ln
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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
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)
由题意得:y的方程式可以写为y=[1/根号下(1-x)]*[ln根号下(1+x)]求导y'=[1/根号下(1-x)]'*[ln根号下(1+x)]+[1/根号下(1-x)]*[ln根号下(1+x)]=[
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(
y=(1+x)⁻¹ln(1-x)y'=(-1)(1+x)⁻²ln(1-x)+(1+x)⁻¹(1-x)⁻¹(-1)
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)/(ͦ
ln(x-y)/ln(x+y)=ln(x-y-x-y)=ln(-2y)
答案在插图:
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)]'=-
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/
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