ln(1 x^2 y^2)d
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分子分母同乘以√x^2+1-x再问:哪里来的分子分母?我问的是第一步是怎么来的?再答:把x+√x^2+1看成(x+√x^2+1)/1,分母看成1
y'=1-2x/(1+x²)=(1+x²-2x)/(1+x²)=(x-1)²/(1+x²)显然y'>0所以y单调增加
再问:极径r积分区域为什么是0
分子有理化,分子分母同乘以-x-√(x²-a²)结果是2lna-ln(-x-√(x²-a²))
答:设极坐标x=cosθ,y=sinθ,1
y'=[ln(x+√(1+x²))]'=1/(x+√(1+x²))*[x+√(1+x²)]'=1/(x+√(1+x²))*[1+2x/2√(1+x²)
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))
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-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)
d(ln(x^2+y))=[1/(x^2+y)].(2xdx+dy)再问:那d(2y-t*y^2)怎么算再答:t是常数d(2y-t*y^2)=(2-2ty)dyt是变数d(2y-t*y^2)=2dy-
y'=(1+x/√(1+x^2))/(x+√(1+x^2))=1/√(1+x^2)y''=-x/(1+x^2)^(3/2)
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)