y+arctanx与y+x交点
@农瑶178：y等于arctanx与y等于x图像有交点吗? -
满肩17836226717…… 唯一交点(0,0) x>0,x>arctanx x
@农瑶178：求通解(1+x^2)y'+y=arctanx -
满肩17836226717…… 解:∵(1+x^2)y'+y=arctanx ==>[(1+x^2)y'+y]e^(arctanx)/(1+x^2)=arctanx*e^(arctanx)/(1+x^2) (等式两端同乘e^(arctanx)/(1+x^2)) ==>e^(arctanx)dy+ye^(arctanx)dx/(1+x^2)=arctanx*e^(arctanx)dx/(1+x^2) ==>e^(arctanx)dy+yd(e^(...
@农瑶178：y=x+arctanx的两条斜渐近线 -
满肩17836226717…… k=lim(x→∞)y/x =1+lim(x→∞)arctanx/x =1+0 =1 b1=lim(x→+∞)(y-x) =lim(x→+∞)arctanx =π/2 b2=lim(x→-∞)(y-x) =lim(x→-∞)arctanx =-π/2 所以,有两条斜渐近线 y=x±π/2
@农瑶178：y=(1+x^2)^arctanx 对y求导 -
满肩17836226717…… y =(1+x^2)^arcatanx lny = arctanx. ln(1+x^2)(1/y)y' = 2xarctanx/(1+x^2) + ln(1+x^2)/(1+x^2) y' = [2xarcanx+ln(1+x^2)]/(1+x^2) . (1+x^2)^arctanx
@农瑶178：证明arctanx+arctany=arctan(x+y/1 - xy),其中xy不等於1 -
满肩17836226717…… 左右2边取正切,左边=(X+Y)/(1-XY)=右边. 左边=arctan[(X+Y)/(1-XY)+Z]/[1-(X+Y)Z/(1-XY)]=arctanc(X+Y+Z-XYZ)/[1-XY-(X+Y)Z]=0 所以X+Y+Z-XYZ=0,既x+y+z=xyz
@农瑶178：tan(arctanX+arctanY)=(X+Y)/(1 - XY) -
满肩17836226717…… ❶证明:tan(arctanX+arctanY)=(X+Y)/(1-XY) 证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tanarctanx)(tanarctany)] =(x+y)/(1-xy)=右边. 【如果看不明白,可令arctanx=α, arctany=β,那么tanα=tanarctanx=x,tanβ=tanarctany=y ❷证明sin(2...
@农瑶178：设y=arctanx+x^y,求dy -
满肩17836226717…… 1 + x^是不是1+x^2? 如果是则 y'=(1+x^2)'*arctanx+(1+x^2)(arctanx)' =2x*arctanx+(1+x^2)*1/(1+x^2) =2x*arctanx+1 所以dy=(2x*arctanx+1)dx
@农瑶178：y'=(1+y^2)/(1+x^2)的通解 -
满肩17836226717…… dy/dx=(1+y^2)/(1+x^2) dy/(1+y^2)=dx/(1+x^2) arctany=arctanx+arctanC y=tan(arctanx+arctanC) =(x+C)/(1-Cx)
@农瑶178：Z=e^xy+(x^2+y^2)arctanx偏导数? -
满肩17836226717…… Z'x=e^(xy) · y + 2x arctanx +(x²+y²) · 1/(1+x²)=ye^(xy) +2xarctanx +(x²+y²)/(1+x²) Z'y=e^(xy) · x +2y arctanx =xe^(xy) +2yarctanx
@农瑶178：设函数 y = ( 1 + X^ )arctanx , 求dy . -
满肩17836226717…… 1 + x^是不是1+x^2?如果是则y'=(1+x^2)'*arctanx+(1+x^2)(arctanx)'=2x*arctanx+(1+x^2)*1/(1+x^2)=2x*arctanx+1所以dy=(2x*arctanx+1)dx
满肩17836226717…… 唯一交点(0,0) x>0,x>arctanx x
@农瑶178：求通解(1+x^2)y'+y=arctanx -
满肩17836226717…… 解:∵(1+x^2)y'+y=arctanx ==>[(1+x^2)y'+y]e^(arctanx)/(1+x^2)=arctanx*e^(arctanx)/(1+x^2) (等式两端同乘e^(arctanx)/(1+x^2)) ==>e^(arctanx)dy+ye^(arctanx)dx/(1+x^2)=arctanx*e^(arctanx)dx/(1+x^2) ==>e^(arctanx)dy+yd(e^(...
@农瑶178：y=x+arctanx的两条斜渐近线 -
满肩17836226717…… k=lim(x→∞)y/x =1+lim(x→∞)arctanx/x =1+0 =1 b1=lim(x→+∞)(y-x) =lim(x→+∞)arctanx =π/2 b2=lim(x→-∞)(y-x) =lim(x→-∞)arctanx =-π/2 所以,有两条斜渐近线 y=x±π/2
@农瑶178：y=(1+x^2)^arctanx 对y求导 -
满肩17836226717…… y =(1+x^2)^arcatanx lny = arctanx. ln(1+x^2)(1/y)y' = 2xarctanx/(1+x^2) + ln(1+x^2)/(1+x^2) y' = [2xarcanx+ln(1+x^2)]/(1+x^2) . (1+x^2)^arctanx
@农瑶178：证明arctanx+arctany=arctan(x+y/1 - xy),其中xy不等於1 -
满肩17836226717…… 左右2边取正切,左边=(X+Y)/(1-XY)=右边. 左边=arctan[(X+Y)/(1-XY)+Z]/[1-(X+Y)Z/(1-XY)]=arctanc(X+Y+Z-XYZ)/[1-XY-(X+Y)Z]=0 所以X+Y+Z-XYZ=0,既x+y+z=xyz
@农瑶178：tan(arctanX+arctanY)=(X+Y)/(1 - XY) -
满肩17836226717…… ❶证明:tan(arctanX+arctanY)=(X+Y)/(1-XY) 证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tanarctanx)(tanarctany)] =(x+y)/(1-xy)=右边. 【如果看不明白,可令arctanx=α, arctany=β,那么tanα=tanarctanx=x,tanβ=tanarctany=y ❷证明sin(2...
@农瑶178：设y=arctanx+x^y,求dy -
满肩17836226717…… 1 + x^是不是1+x^2? 如果是则 y'=(1+x^2)'*arctanx+(1+x^2)(arctanx)' =2x*arctanx+(1+x^2)*1/(1+x^2) =2x*arctanx+1 所以dy=(2x*arctanx+1)dx
@农瑶178：y'=(1+y^2)/(1+x^2)的通解 -
满肩17836226717…… dy/dx=(1+y^2)/(1+x^2) dy/(1+y^2)=dx/(1+x^2) arctany=arctanx+arctanC y=tan(arctanx+arctanC) =(x+C)/(1-Cx)
@农瑶178：Z=e^xy+(x^2+y^2)arctanx偏导数? -
满肩17836226717…… Z'x=e^(xy) · y + 2x arctanx +(x²+y²) · 1/(1+x²)=ye^(xy) +2xarctanx +(x²+y²)/(1+x²) Z'y=e^(xy) · x +2y arctanx =xe^(xy) +2yarctanx
@农瑶178：设函数 y = ( 1 + X^ )arctanx , 求dy . -
满肩17836226717…… 1 + x^是不是1+x^2?如果是则y'=(1+x^2)'*arctanx+(1+x^2)(arctanx)'=2x*arctanx+(1+x^2)*1/(1+x^2)=2x*arctanx+1所以dy=(2x*arctanx+1)dx
