z=x分之y,而x=e的t次方,y=1-e的2t次方,求dz÷dt
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![z=x分之y,而x=e的t次方,y=1-e的2t次方,求dz÷dt](/uploads/image/f/921737-65-7.jpg?t=z%3Dx%E5%88%86%E4%B9%8By%2C%E8%80%8Cx%3De%E7%9A%84t%E6%AC%A1%E6%96%B9%2Cy%3D1-e%E7%9A%842t%E6%AC%A1%E6%96%B9%2C%E6%B1%82dz%C3%B7dt)
左边=(9^x*10^y*16^z)/(8^x*9^y*15^z)=3^(2x-2y-z)*(2^y+4z-3x)*5^(y-z)=2^1所以2x-2y-z=0y-4z-3x=1y-z=0x=-1/5
y+z分之x=x+z分之y=x+y分之z;合比定理:x/(x+y+z)=y/(x+y+z)=z/(x+y+z);得x=y=z,y+z分之x的值=1/2
设(x+y-z)/z=(x-y+z)/y=(-x+y+z)/x=k则(1)x+y-z=kz(2)x-y+z=ky(3)-x+y+z=kx(1)+(2)+(3)得x+y+z=k(x+y+z)∴k=1时,
(太麻烦拉,给点分啊!)设v=x*x-y*y,u=exp{xy}那么dv/dx=2x(这里应该用偏导符号,代替一下),dv/dy=2y,du/dx=y*exp{xy},du/dy=x*exp{xy}那
x和y换一下不就得到:2x=(e的y次方减e的负y次方)设e的y次方等于t所以t+1/t=2x,就是t2-2tx+1=0解得:t=x加减根号下(x2-1)因为原函数的值域就是反函数的定义域2分之e的x
z'x=2e^(2x+y)z'y=e^(2x+y)所以dz=2e^(2x+y)dx+e^(2x+y)dy
同时取对数令:xlg3=ylg4=zlg6=m则1/x=lg3/m1/y=lg4/m1/z=lg6/m1/z-1/x=(lg6-lg3)/m=lg2/m1/(2y)=(1/2)lg4/m=lg2/m得
求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(
你想说这个问题?z=e^(x^2+2xy)应该是y=e^(x^2+2xy)(2x+2y)i+e^(x^2+2xy)2xj
令(y+z)/x=(z+x)/y=(x+y)/z=ky+z=kxx+z=kyx+y=kz2(x+y+z)=k(x+y+z)2(x+y+z)=k(x+y+z)(2-k)(x+y+z)=0(x+y+z≠0
z=arctan(x*e^x)z'={1/[1+(x*e^x)^2]}*(x*e^x)'(x*e^x)'=x'*e^x+x*(e^x)'=e^x+x*e^x=(x+1)*e^x所以dz/dx=(x+1
z'x=yx^(y-1),z'y=x^ylnxx't=e^t,y't=1dz/dt=z'x*x't+z'y*y't=yx^(y-1)e^t+x^ylnx再问:最后答案是dz/dt=2te^(t^2),
已知(xy分之x+y)的-1次方=1,则xy=x+yy=x/(x-1)(yz分之y+z)的-1次方=2,则yz=2y+2zz=2y/(y-2)(xz分之x+z)的-1次方=3,则xz=3x+3zz=3
(y+z)/x=(z+x)/y=(x+y)/z=k所以:y+z=kx,z+x=ky,x+y=kz上述三个等式相加,得到:2(x+y+z)=k(x+y+z)得到:k=2
(x+y-z)/z=(y+z-x)/x=(z+x-y)/y[x+y]/z-1=[y+z]/x-1=[z+x]/y-1[x+y]/z=[y+z]/x=[z+x]/y设[x+y]/z=[y+z]/x=[z
求二元函数全微分z=f[x²-y²,e^(xy)]设z=f(u,v),u=x²-y²,v=e^(xy)则dz=(∂f/∂u)du+(
e^(-xy)-x^2*y+e^z=z,令F(x,y,z)=e^(-xy)-x^2*y+e^z-z=0分别对F取x,y,z的偏导数,可得əF/əx=e^(-xy)*(-y)-2xy
设2分之x+y=3分之y+z=5分之x+z=k则x+y=2k,y+z=3k,x+z=5k3式相加得2(x+y+z)=10k=18*2=36,k=3.6,x+y+z=5kz=5k-2k=3k=10.8x