x=ln(1 t^2),y=arctant,求dy dx
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y'=(lnlnx)'/lnlnx=(lnx)'/lnxlnlnx=1/xlnxlnlnx
y'=[1/(1+x^2)]*(1+x^2)'=[1/(1+x^2)]*2x=2x/(1+x^2)
chainruley=f(g(x))y'=g'(x)f'(g(x))
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(
表示以e为底的对数函数符号
dy/dx=[1-1/(1+t²)]/[2t/(1+t²)]=t/2d²y/dx²=(1/2)*dt/dx=(1/2)/(dx/dt)=(1/2)/[2t/(1
y=ln(1+t)t=e^y-1x=e^(2y)-e^y两边同时对x求导得dy/dx=1/(2e^(2y)-e^y)=1/(2(1+t)^2-1+t)=1/(2t^2+3t+1)
Y=[LN(1-X)]^2?Y'=2LN|1-X|/(1-X)(-1)=-2LN|1-X|/(1-X)
y=ln(1-x^2)y'=(1-x^2)'/(1-x^2)=-2x/(1-x^2)
明显你是对的.答案是哪里来的,明显不对.
2x/(1+x^2)
y'=ln(2x^-1)'=(x/2)*2*(-1)/x^2=-1/x
x≤0时√x^2=-x所以y=0x>0时√x^2=x所以y=ln(2x+1)
先分别求出dx/dt和dy/dt,假设A=dx/dt,B=dy/dt然后用B/A得出dy/dx设C=B/A=dy/dxC中只含有t.因此,d^2y/dx^2=C/dt乘以dx/dt的倒数(dt/dx)
分别算出dx,dy,然后相除就行详见参考资料
dx/dt=2t/(1+t²)dy/dt=1/(1+t²)dy/dx=1/(2t)d(dx/dt)/dt=(2-4t²)/(1+t²)²d(dy/dt
x=tany+ln(cosy^2),dy/dx=(dx/dy)^-1=(tany-1)^-2,y"=d(dy/dx)/dy*dy/dx=-2secy^2/(tany-1)^5