y=f(x)由方程x-y ½siny=0所确定的隐函数,求d²y dx²
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![y=f(x)由方程x-y ½siny=0所确定的隐函数,求d²y dx²](/uploads/image/f/910084-4-4.jpg?t=y%3Df%28x%29%E7%94%B1%E6%96%B9%E7%A8%8Bx-y+%C2%BDsiny%3D0%E6%89%80%E7%A1%AE%E5%AE%9A%E7%9A%84%E9%9A%90%E5%87%BD%E6%95%B0%2C%E6%B1%82d%C2%B2y+dx%C2%B2)
两边对x求导:2cos(x^2+y)*(-sin(x^2+y))*(2x+y')=1所以y'=-1/sin(2x^2+2y)-2x再问:求f'(x)```再答:y'就是f'(x)啊。。。。。
两边都对x求导有(2x+dy/dx)/(xˆ2+y)=3xˆ2y+xˆ3dy/dx+cosx得dy/dx=(3xˆ4y+3xˆ2yˆ2+x&
两边对x求导有y'e^y=y+xy'整理解得y‘=dy/dx=x/(e^y-x)
方程两边分别对x求导得:y'e^y-1-2x+2y'=0移项得:(e^y+2)y'=2x+1所以:y'=dy/dx=(2x+1)/(e^y+2)
y'=cos(x+y)(1+y')y'=cos(x+y)/(1-cos(x+y))
两边对x求导:y'e^y+(1+y')cos(x+y)=0,1)这里可得到y'=-cos(x+y)/[e^y+cos(x+y)]再对1)求导:y"e^y+(y')^2e^y+y"cos(x+y)-(1
令u=x+y则y=f(u)两边对x求导,得:y'=f'(u)*u'=f'(u)*(1+y')解得:y'=f'(u)/[1-f'(u)]故dy=f'(x+y)/[1-f'(x+y)]*dx
xy+y^2-2x=0y+xy'+2yy'-2=0(x+2y)y'=2-yy'=(2-y)/(x+2y)dy/dx=(2-y)/(x+2y)
再问:是否还能给出一种利用题目所给的条件(关于x,y,z的函数)去证明的方法吗?再答:这就是课本上隐函数求导公式的应用,你想得太多了,没有必要的!
两端对x求导数(把y看作x的函数),则1-y'=e^(xy)*(1*y+x*y')y'[xe^(xy)+1]=1-ye^(xy)dy/dx=y'=[1-ye^(xy)]/[xe^(xy)+1]
xe^f(y)=ln2009e^ye^f(y)+xe^f(y)*f'(y)*y'=y'e^f(y)(1+xf'y')=y'e^f*f'*y
两边对x求导得:2yy'*f(x)+y^2f'(x)+f(x)+xf'(x)=2x得:y'=[2x-xf'(x)-y^2f'(x)]/(2yf(x)]dy=[2x-xf'(x)-y^2f'(x)]/(
y'=(x)'e^y+x(e^y)'y'=e^y+xe^y*y'再问:x(e^y)'=xe^y*y'?再答:对,因为y是x的函数,根据复合函数求导法,可得
两边对x求导xy^2+sinx=e^yy^2+2xyy'+cosx=e^y*y'y'(e^y-2xy)=y^2+cosxy'=(y^2+cosx)/(e^y-2xy)
F(x,y)=x^2+y^2-ln(x+2y)Fx=2x-1/(x+2y)Fy=2y-2/(x+2y)F(x)=-Fx/Fy=-[2x(x+2y)-1]/[2y(x+2y)-2]
两边对x求导:1+y'=y'e^y得dy/dx=y'=1/(e^y-1)
两边对x求导:3x^2-3y^2-6xyy'+6y^2y'=0得y'=(y^2-x^2)/[2(y^2-xy)]=(y+x)/(2y)令y'=0,得y+x=0,将y=-x代入原方程:x^3-3x^3-