求函数y=cos∧3e∧x的微分dy
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我来试试吧...z=e^xy*cos(x+y)Z'x=ye^xycos(x+y)-e^xysin(x+y)Z'y=xe^xycos(x+y)-e^xysin(x+y)故dZ=[ye^xycos(x+y
y=2cos(x+π4)cos(x−π4)+3sin2x=2(12cos2x−12sin2x)+3sin2x=cos2x+3sin2x=2sin(2x+π6)∴函数y=2cos(x+π4)cos(x−
y=sin∧4x+2√3sinxcosx-cos∧4x(3前面是根号,不是积分?)=sin∧4x-cos∧4x+√3sin2x=(sin²x-cos²x)(sin²x+c
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两边微分-sin(x+y)(dx+dy)+e^y*dy=0[e^y-sin(x+y)]dy=sin(x+y)dxdy=sin(x+y)dx/[e^y-sin(x+y)]
微分的乘积法则和链式法则学过吗?这两个都是微分基本的法则,做这道题时都会用到.y=e^(-x)cos(3-x)dy/dx={d[e^(-x)]/dx}cos(3-x)+e^(-x){d[cos(3-x
(1)y'=e^arctan√x(1/1+x)(1/2x的(-1/2次方))(2)y‘=1/cose^x(-sine^x)e^x
两边求导,-sin(x+y)(1+y`)+e^yy`=1,dy=1+sin(y+x)/e^y-sin(x+y)dx再问:亲,这是正确的么?我是帮人问的==对的就给分了啊!
f(x,y)=e^(x+y)+cos(xy)=0 //: 利用隐函数存在定理:f 'x(x,y)=e^
微分dy=[-e的(-x)cos(3-x)+e的(-x)sin(3-x)]dx要想求导的话就直接把dx移到前边去就好了.
1)y'=2/[(2x+1)ln3]2)y'=2e^(2x)cos(3x+1)-3e^(2x)sin(3x+1)
y=cos(π/3-x)y'=-sin(π/3-x)*(-1)=sin(π/3-x)y=e^3xy'=e^(3x)*3=3e^(3x)y=In(3-x)y'=1/(3-x)*(-1)=1/(x-3)y
隐函数直接求导数.e^(x+y)(1+y')-sin(xy)(y+xy')=0解出y'即为dy/dx=[e^(x+y)-y*sin(xy)]/[e^(x+y)-x*sin(xy)]
f'(x)=e^x+4>0所以,f(x)在R上是递增,f(0)=-2
y=cos²x-3sinx=1-sin²x-3sinx=-(sin²x+3sinx)+1=-(sin²x+3sinx+9/4)+13/4=-(sinx+3/2)
y=cos^2x-sinx=1-sin²x-sinx=-(sinx+1/2)²+5/4所以当sinx=-1/2时,有最大值=5/4当sinx=1时,有最小值=1-1-1=-1值域为
y=3cosX-cos(2X)=3cosx-(2*(cosx^)2-1)=-2(cosx)^2+3cosx+1=-2(cosx-3/4)^2+17/8当cos=3/4时,y有最大值,为17/8当cos
y'=-sin(4-3X)*(-3)=3sin(4-3X)
求导,令一介导等0;y=e^x-yx^2-2y^3;两边对x求导得y'=e^x-y'x^2-2xy-6y^2y';令y'=0得2xy=e^x
y=y(x)不能找到显形式,只能表为f(x,y)=0即原题中给出的形式dy/dx计算过程与结果如下图