设y=ln(arctan(1-2)),求dy-搜答案-神马作文网

y=ln(1+x)y′=1/(1+x)y′′=-1/(1+x)²y′′′=(-1)(-2)[1/(1+x)³].y^n=(-1)(-2)...(-n+1)[1/(1+x)^n]

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

点击放大,有详细过程.

arctan(y)=x+1, y=?

两边取正切y=tan(x+1)

占座马上回答再答：

symsx;y=atan((x^2-1)^(1/2))-log(x)/((x^2-1)^(1/2))y=atan((x^2-1)^(1/2))-log(x)/(x^2-1)^(1/2)>>diff(y

设y=ln 1/x +ln2 求y'

两种方法：1.求ln1/x的导数时,结果是1/(1/x)=x,因为是复合函数,此时还要乘以1/x的导数,即-1/x^2,最后结果是-1/x,ln2是常数,导数是0所以y'=-1/x;2.如果你上面的方

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y=ln(1＋x)y'=1/(1＋x)y''=－1/(1＋x)²熟记求导公式

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

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

y=2^arccot(x)-sin3y'=2^arccotx*[-1/(1+x²)]*ln2dy=2^arccotx*[-1/(1+x²)]*ln2dx

y=f{g[h(p(x))]}y'=f'(g)g'(h)h'(p)p'(x)y'=1/cos(arctan(sinx))*(-sin(arctan(sinx))*cosx/(1+sinx^2)=-ta

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