求函数y=e∧sin²(1-x)的导数
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y=sin(e^-x)这是一个复合函数=cos(e^-x)*(e^-x)‘=-e^-x*cos(e^-x)
求函数导数y=e^2x*sin^2xy′=2e^(2x)sin²x+2e^(2x)sinxcosx;很高兴为您解答,skyhunter002为您答疑解惑如果本题有什么不明白可以追问,
y=e^x/(x+1)y'=[e^x(x+1)-e^x]/(x+1)^2=xe^x/(1+x)^2
y=sin∧2(2x-1)的导数y'=2sin(2x-1)*cos(2x-1)*2=2sin(4x-2)
1)x=0代入方程:1-e^y=0,得y(0)=0两边对X求导:e^x-y'e^y=cos(xy)(y+xy')y'=[e^x-ycos(xy)]/[xcos(xy)+e^y]代入x=0,y(0)=0
再答:隐函数高阶求导。再答:
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
e^y=sin(x+y)两边求导得e^y*y'=cos(x+y)(x+y)'=cos(x+y)(1+y')=cos(x+y)+y'cos(x+y)[e^y-cos(x+y)]y'=cos(x+y)y'
f'(x)=lim(h→0)[f(x+h)-f(x)]/h=lim(h→0)[sin(e^(x+h)+1)-sin(e^x+1)]/h由和差化积公式得=lim(h→0)2cos[(e^(x+h)+e^
不知道sinx是指数还是其他的,如果是y=xsinx的话,f'(x)=sinx+xcosx,如果sinx是x的指数的话,f'(x)=(sinx-1)*x^(sinx-1)
y=sin(e∧x+1)dy=cos(e^x+1)d(e^x+1)=e^xcos(e^x+1)dx
x+3>0,且ln(x+3)≠0得:x>-3且x≠-2所以,定义域为(-3,-2)U(-2,+∞)
因为你没给初值,所以我只能给你一簇图像,里面的任意一条曲线都满足该微分方程.[x,y]=meshgrid(linspace(-sqrt(2*pi),sqrt(2*pi)));streamslice(x
dy=[(cosx-sinx)/e^x]*dx
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
令A=sinx,则y=A^2+A-1,{A|-1
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)]=[