设 y=sin²(3x 5),求y
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![设 y=sin²(3x 5),求y](/uploads/image/f/7243559-71-9.jpg?t=%E8%AE%BE+y%3Dsin%C2%B2%283x+5%29%2C%E6%B1%82y)
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=2cosxdcosx-5x^4dx=-2sinxcosxdx-5x^4dx=(-sin2x-5x^4)dx
∵(π3+4x)+(π6-4x)=π2,∴cos(4x-π6)=cos(π6-4x)=sin(π3+4x),∴原式就是y=2sin(4x+π3),这个函数的最小正周期为2π4,即T=π2.当-π2+2
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
cos(x+y)(1+y')=y+xy'dy/dx=y'=[y-cos(x+y)]/[cos(x+y)-x]
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
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
z=sin(x²y²)+3x-5y²+1所以δz/δx=cos(x²y²)*2xy²+3δz/δy=cos(x²y²)*
dy/dx=-2cosxsinx-5x的4次方所以dy=(-sin2x-5x的4次方)dx
1=x?
函数y=3sin(2x+φ)的对称轴是x=-φ/2+kπ/2,k是整数由π/3=-φ/2+π/4+kπ/2,0
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)]=[