x^2ln(1 x)arctanx f(n)(0)
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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)
极限是0吗?因为x->0,ln(1+x)与x是等价无穷小,因为lim(ln(1+x)/x))=1,所以ln(1+x)极限为0,而1/x->正无穷大,所以arctan1/x的极限为pai/2,二者相乘为
(1)y'=e^arctan√x(1/1+x)(1/2x的(-1/2次方))(2)y‘=1/cose^x(-sine^x)e^x
再问:л�˰�再问:��
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
你的第一题的变形没错啊,你将化出的式子中e^x换成t,e^2x换成t^2就可看到结果!至于第二题,你应该提高你计算的准确率
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
∫arctan(1+√x)dx令√x=tx=t^2dx=dt^2原式化为∫arctan(1+t)*dt^2=t^2arctan(1+t)-∫t^2*1/(1+t^2)dt=t^2arctan(1+t)
因为x=(x^1/2)^2那么dx=2d(x^1/2)所以原式=2arctan(x^1/2)d(x^1/2)=2/[1+(x^1/2)^2
对x求导1/√(x²+y²)*[1/2√(x²+y²)]*(2x+2y*y')=1/(1+y²/x²)]*(y'*x-y)/x²(
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
y'=1/[1+(x^2+1)^2]×(x^2+1)'=2x/(x^4+2x^2+2)再问:
tan(A+B)=(tanA+tanB)/(1-tanAtanB)tan(arctan(1-x)+arctan(1+x))=(1-x+1+x)/(1-(1-x)(1+x))=2/x^2arctan(1