(1 x^2)dy=arctanxdx
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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)
首先由格林公式得∮Pdx+Qdy=∫∫(Q'(x)-P'(y))dxdy然后化为极坐标的形式积分就可以出来了!我也是新手,一些数学符号弄不出来,希望你能看懂,当然高数的内容还是要多看课本,仔细比较,多
x^2+y^2-e^(arctan(y/x))=02x+2y*y'-(arctan(y/x))'e^(arctan(y/x))=02x+2y*y'-1/(1+y^2/x^2)*(y'x-y)/x^2*
dz/dx=arctan(xy)+xy/[1+(xy)^2](dz/dx)|(1,1)=π/4+1/2(dz/dy)|(1,1)=x^2/[1+(xy)^2]=1/2
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
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
dx/dt=1+1/(t²+1)+0=(t²+2)/(t²+1)dy/dt=3t²+6所以dy/dx=(dy/dt)/(dx/dt)=(3t²+6)/
arctanx'=1/(1+x^2)y=arctan(x+1)^1/2y'=1/(1+(x+1)^1/2^2)*(x+1)^1/2'y'=1/(x+2)*1/2(x+1)^(-1/2)y'=1/[2(
y=2^arccot(x)-sin3y'=2^arccotx*[-1/(1+x²)]*ln2dy=2^arccotx*[-1/(1+x²)]*ln2dx
dy/dx=1/[1+(1+x^2)]*2x刚考过导数表示非常苦逼.哎我还是讲清楚点这是复合函数,把它拆成y=arctanuu=1+x^2再分别求导数再问:·再答:==dy/dx=[arctan(1+
dy/dx=[arctan(1+x^2)]'*(1+x^2)'=1/[1+(1+x^2)^2]*2x话说,我刚回答了一道一样的.再问:下面一行就是最终过程了??不好意思,因为我是数学白痴再答:嗯。。导
y'=(4+x^2)'(arctanx/2)+(4+x^2)(arctanx/2)'=2x(arctanx/2)+(4+x²)*[1/(1+x²/4)]*1/2=2x(arctan
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:
这不就是求导数吗dy/dx=(1/a)*(1/(1+x^2)*(1/a)=1/[a^2*(1+x^2)]
此题复合求导dy=d[arctan(1-x/1+x)]=[1/(1+(1-x/1+x)^2)]·(1-x/1+x)';注:(arctanx)'=1/(1+x^2)=-(1/(x^2+1))