设ln√x² y²=arctany x-搜答案-智慧作业帮

两边对【x】求导,注意,y是x的函数,利用复合函数求导1/[1+(y/x)^2]×(y/x)'=1/2×1/(x^2+y^2)×(x^2+y^2)',也就是：x^2/(x^2+y^2)×(xy'-y)

设y=ln ln ln x,求y’

y'=(lnlnx)'/lnlnx=(lnx)'/lnxlnlnx=1/xlnxlnlnx

占座马上回答再答：

（1）y'=e^arctan√x（1/1+x）(1/2x的（-1/2次方）)(2)y‘=1/cose^x(-sine^x)e^x

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y=arctan(a/x)+1/2[ln(x-a)-ln(x+a)],利用复合函数求导的链锁规则,有y'=1/(1+(a/x)^2)*(-a/x^2)+1/2[1/(x-a)]-1/(x+a)]=-a

dy/dx=(y-2x)/(2y-x),要详解吗?再问：��д

两边同时对x求导,得(2x+2yy')/(x²＋y²)=1/(1+y²/x²)·（xy'-y）/x²(2x+2yy')/(x²＋y²

两边同时求导根据链式法则1/2（x²+y²）’/(x²+y²)=(x/y)'/[1+(x/y)²]1/2(2x+2yy')/(x²+y

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

两边求导(y'x-y/x^2)/[1+(y/x)^2]=x+yy'/(x^2+y^2)^1/2整理y'x-y=(x+yy')(x^2+y^2)^1/2

左右两边对x求导,注意y是关于x的复合函数：(x^2+y^2)^(-1/2)*(2x+2y*y')=[1/(1+y/x)]*(y/x)^(-1/2)*[(y'x-y)/(x^2)]把y'归在一边,就可

对x求导1/√(x²+y²)*[1/2√(x²+y²)]*(2x+2y*y')=1/(1+y²/x²)]*(y'*x-y)/x²(

对x求导0.5*1/(x²+y²)*(x²+y²)'=1/[1+(y/x)²]*(y/x)'0.5/(x²+y²)*(2x+2y*

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

即0.5ln(x^2+y^2)=arctan(y/x)对x求导得到0.5(2x+2y*y')/(x^2+y^2)=1/(1+y^2/x^2)*(y/x)'即(2x+2y*y')/(x^2+y^2)=2

原式化简为1/2ln(x^2+y^2)=arctany/x两边对x求导,得1/2×1/(x^2+y^2)×(2x+2yy')=1/[1+(y/x)^2]×(y'x-y)/x^2化简得y'=(x+y)/