求隐函数y的二阶导数2arctany x=ln(x^2 y^2)
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![求隐函数y的二阶导数2arctany x=ln(x^2 y^2)](/uploads/image/f/5750084-20-4.jpg?t=%E6%B1%82%E9%9A%90%E5%87%BD%E6%95%B0y%E7%9A%84%E4%BA%8C%E9%98%B6%E5%AF%BC%E6%95%B02arctany+x%3Dln%28x%5E2+y%5E2%29)
/>y=ln(1+x^2)y'=2x/(1+x^2)y''=[2(1+x^2)-2x(2x)]/[(1+x^2)^2]=(2+2x^2-4x^2)/[(1+x^2)^2]=2(1-2x^2)/[(1+
y'=2xf'(x^2)y''=2f'(x^2)+4x^2f''(x^2)
y=x^2lnx所以y‘=(x^2)'lnx+x^2(lnx)'=2xlnx+x^2*1/x=2xlnx+xy''=(2xlnx+x)'=2lnx+2x*(lnx)'+1=2lnx+2x*1/x+1=
-sinx+4e的2x方再答:y=-sinx+4e∧2x再问:大哥给个过程呗,考试用再答:等一下再答:y'=cosx+2e∧2xy''=-sinx+4e∧2x再答:那个2x也要求导的所以会这样
-y/x^2
y’=k(-5x*sin5x+cos5x);y’’=k[(-5x*5cosx-5sin5x)–5sin5x]=k(-10sin5x-25xcos5x);代入微分方程得:k(-10sin5x-25xco
xy³=y+xy³+3xy²*y'=y'+1(3xy²-1)y'=1-y³y'=(1-y³)/(3xy²-1)3y²y'
不够明白,是这样吗:
x-y+1/2siny=0两边对x求导得1-y'+1/2cosy*y'=0y'=2/(2-cosy)y''=dy'/dx=(dy'/dy)*(dy/dx)=[-2/(2-cosy)²]*si
y=xf(u),u=x^2,u'=2xy'=f(u)+xf'(u)u'=f(u)+2x^2f'(u)y"=f'(u)+4xf'(u)+2x^2f"(u)u'=f'(u)+4xf'(u)+4x^3f"(
(1)y=f(x)d^2y/dx^2=d(f'(x))/dx=f''(x)(2)y=ln[f(x)]dy/dx=f'(x)/f(x)d^2y/dx^2=d[f'(x)/f(x)]/dx=[f''(x)
y-1=xe^y两边同时对x求导得y'=e^y+xe^y*y'(1-xe^y)y'=e^yy'=e^y/(1-xe^y)=e^y/(2-y)y''=(e^y*y'+e^y*y')/(2-y)²
两边关于x求导,注意y是x的函数y'cosy=[1/(x+y)]*(1+y').①解得y'=1/(x+y)÷[cosy-1/(x+y)].②对①两边关于x求导可得y''cosy-(y')²s
1.y'=x^2(2^x)'+(2^x)*2x=x^2*2^x*ln2+(2^x)*2xy''=(x^2*2^x*ln2+(2^x)*2x)*ln2+2x(2^x)ln2+2^x*22.y'=e^xc
关于x求导得:2x-2yy′=0y′=x÷yy′′=(y-xy′)÷y^2=(y^2-x^2)÷y^3=−4÷y^3
两边对x求导,则3x^2+3y^2*y'-(y+xy')=0(1)所以,y'=(y-3x^2)/(3y^2-x)(2)(1)两端对x继续求导,则6x+6y*(y')^2+3y^2*y''-(y'+y'
一阶导2x-2yy'=0y'=x/y二阶导y''=(y-xy')/y^2y''=(y-x^2/y)/y^2=(y^2-x^2)/y^3=-4/y^3不知道这么算对不对?好久以前学的有点忘记了
两边对x求导得:y'=e^y+xy'e^yy'=e^y/(1-xe^y)y''=dy'/dx=[y'e^y(1-xe^y)-(-e^y-xy'e^y)e^y]/(1-xe^y)²=(2-x)
y=1+xe^y方程两边求导y'=e^y+xe^y*y'y'(1-xe^y)=e^yy'=(e^y)/(1-xe^y)y''={e^y*y'*(1-xe^y)+e^y[e^y+xe^y*y']}/(1