arctany=t 则y=tant
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两边对x求导得1/[1+(y/x)^2]*(y/x)'=1/[ln(x^2+y^2)]*[ln(x^2+y^2)]'1/[1+(y/x)^2]*(y'x-y)/x^2=1/[2ln(x^2+y^2)]
F(x,y)=A(B+arctanx/2)(C+arctany/3)F(-∞,-∞)=A(B-π/2)(C-π/2)=0F(-∞,+∞)=A(B-π/2)(C+π/2)=0F(+∞,-∞)=A(B+π
第一题上面已有朋友回答第二题可以先化简得:y'=y^2\(-x+2xy+y^2),也可记为dy\dx=y^2\(-x+2xy+y^2),则dx\dy=(-x+2xy+y^2)\y^2,化简得:dx\d
1/2*ln(x^2+y^2)=arctany/x两边对x求导,得1/2*1/(x^2+y^2)*(2x+2y*y')=1/[1+(y/x)^2]*(y'*x-y)/x^2化简得y'=(x+y)/(x
y+arctany-x=0dy/dx+1/(1+y^2)dy/dx-1=0dy/dx(1+1/(1+y^2)=1dy/dx=(1+y^2)/(2+y^2)
一般情况下,ω是正值.如果它是负值,可以先把他变成正数就是了.其实,ω的正负其实不影响他的周期. 例如:y=sin(-|ω|x+θ)+K=-sin(|ω|x-θ)+K,它的周期还是T=2π/ω下面哪
y''=-(2+2y^2)/y
x+arctany=y两边对x求导有:1+y'/(1+y²)=y'整理得:y'=1+1/y²
设y=2arctan(y/x),求dy/dx,d²y/dx².设F(x,y)=y-2arctan(y/x)=0,则dy/dx=-(∂F/∂x)/(ͦ
答案在插图:
见图再问:不好意思啊~题目看错了,题目如图啊~
利用概率分布函数特性F(正无穷,正无穷)=1,F(负无穷,负无穷)=0,带入就是A(B+π/2)(C+π/2)=1A(B-π/2)(C-π/2)=0展开后,两式相加:ABC=1/2-(π^2)/4再问
假设地理纬度为φ,指时针的高度为H,要刻划的时间与正午的差值为T;时间线与指时针的夹角为A,距离为D只是一个三角函数的公式,是用来具体运算,没有太大意义
1.y=arcsin(cosx)y'=[1/√(1-cos²x)](-sinx)=-sinx√(1-cos²x)/sin²x=-|sinx|/sinx∴当sinx>0时y
是(arctany)/x还是arctan(y/x)?如果是z=(arctany)/x,则∂z/∂x=-(arctany)/x²∂z/∂y=1/
利用查表或反函数求导法可求得(arctanu)'=1/(1+u^2)∴上述方程两边分别对x求导可得1+y'=y'/(1+y^2)=>(1+y^2)+(1+y^2)y'=y'=>(1+y^2)+y^2y
tan(2x-y)=[tanx+tan(x-y)]/[1-tanxtan(x-y)]=(1/10)/(1+1/5)=1/12
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
❶证明:tan(arctanX+arctanY)=(X+Y)/(1-XY)证明:tan(arctanx+arctany)=(tanarctanx+tanarctany)/[1-(tana