设y=sin²x 2x,求y
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e'表示对自然对数e求导,e'=0但是在dy/dx的过程中由于分子和分母都有e',可以约掉,所以不用急着把分子分母都等于0,这样就做不出来了.dy/dx=(dy/dt)/(dx/dt)dy/dt=(e
y=sinx/(1+x^2)先求导:y'=cosx*(1+x^2)-sinx*(2x)/(1+x^2)^2那么,dy=y'dx=[cosx*(1+x^2)-sinx*(2x)]dx/(1+x^2)^2
y=sin(x+y),y'=cos(x+y)*(1+y'),y'=cos(x+y)/(1-cos(x+y))=dy/dx
y=sinx²+sin²x∴y'=cos(x²)*(x²)'+2sinx*(sinx)'=2x*cos(x²)+2sinxcosx=2x*cos(x&
再答:隐函数高阶求导。再答:
e^(xy)+sin(xy)=y(y+xy')e^(xy)+(y+xy')cos(xy)=y'y'=(ye^(xy)+ycos(xy))/(1-xe^(xy)-xcos(xy))
dy=dsin(x+y)dy=cos(x+y)d(x+y)dy=cos(x+y)(dx+dy)dy=cos(x+y)dx+cos(x+y)dy所以dy/dx=cos(x+y)/[1-cos(x+y)]
这道题你先看sinx必然大于等于零吧,sin((1-y)x)也必然大于等于零的吧?整个函数都是大于等于0的吧?那么你只要找到可以让函数取到零的x和y就可以得到最小值0那么试着凑一下,y=1,x=pai
∂z/∂x=cos(x-y)∂z/∂y=-cos(x-y)dz=∂z/∂x*dx+∂z/∂y*dy=co
cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]
用辅助角公式~sin的就不用管它外面就只有y再问:能不能写一下详细过程,谢谢再答:不好意思我竞赛没认真读一般一试的填空我都是用猜的我现在高三备高考而且三角的转化我不在行你可以去竞赛吧问问里面的人比较牛
两边同时对x微分得dcos(x^2+y)=dx,即-sin(x^2+y)(2dx+dy)=dx,将dy移过去,变形得到-(1+2sin(x^2+y))dx/sin(x^2+y)=dy
y=sin(xy)dy/dx=cos(xy)*y=ycos(xy)d²y/dx²=-ysin(xy)*y=-y²sin(xy)
若看不清楚,可点击放大.
y=e^(x/2)+x^2*sin√xy′=1/2*e^(x/2)+2x*sin√x+x^2*1/(2√x)*cos√x
原式y=sinx^2+2xdy/dx=2x·cosx^2+2
1=x?
dy/dx=-fx/fy,你自己可以算吧
e^(x+y)+sin(xy)=1e^(x+y)*(1+y')+cos(xy)(y+xy')=0y'*[e*(x+y)+xcos(xy)]=-[ycos(xy)+e^(x+y)]y'=-[ycos(x
化为:e^(ylnx)-e^y=sin(xy)两边对x求导:e^(ylnx)(y'lnx+y/x)-y'e^y=cos(xy)(y+xy')y'[lnxe^(ylnx)-e^y-xcos(xy)]=[