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3y xCosx

y=(1-x)/cosx dy/dx=(1-x)'/cosx+(1-x)(1/cosx)'=-1/cosx+(1-x)(sinx)/cos^2x =(sinx-xsinx-cosx)/cos^2x y=1-(x/cosx) dy/dx=-1/cosx-x*(1/cosx)'=-1/cosx-x(sinx)/cos^2x=-(xsinx+cosx)/cos^2x

解:∵(3-4y-cosx)2+(4+3y+sinx)2=([(3?4y)?cosx]2+[(4+3y)?(?sinx)]2)2,类比两点间的距离公式|AB|=(x1?x2)2+(y1?y2)2,而且3(3-4y)+4(4+3y)-25=0,∴所求的式子为直线3x+4y-25=0上的一点到圆x2+y2=1上的一点的距离的平方,画图可知,过...

猜[2ysin(x/y)+ 3xcos(x/y)]dy-3ycos(x/y)dx=0, 两边都除以y,得[2sin(x/y)+(3x/y)cos(x/y)]dy-3cos(x/y)dx=0, 设x=uy,则dx=ydu+udy,上式变为 [2sinu+3ucosu]dy-2cosu(ydu+udy)=0, 整理得(2sinu+ucosu)dy=2ycosudu, 分离变量得dy/y=2cosudu/(2s...

f(x) = sinx - ∫(0~x) (x - t) f(t) dt = sinx - x∫(0~x) f(t) dt + ∫(0~x) tf(t) dt,之后两边对x求导 f'(x) = cosx - [x' · ∫(0~x) f(t) dt + x · f(x)] + xf(x) f'(x) = cosx - ∫(0~x) f(t) dt,两边再对x求导 f''(x) = - sinx - f(x) ==> y'' ...

1原式=2sin2Acos2A(2-2cosA^2)/cos2A(2-2cos2A^2)=4sin2AsinA^2/2sin2A^2=2sinA^2/sin2A=2sin2A^2/2sinAcosA=tanA 2.乘以cosx^2,2sinx^2-5cosx-i0cosx^2=0,2-5cosx-12cosx^2=0,解出cosx=1/4或-2/3,第二象限,A=arccos-2/3

> with(plots); > plot3d(sin(x)*cos(y), x = -Pi .. Pi, y = -Pi .. Pi, axes = boxed);

解: 令y=cosx/(sinx-3) ysinx-cosx=3y √(y²+1²)sin(x-θ)=3y,(其中,cotθ=y) sin(x-θ)=3y/√(y²+1) -1≤sin(x-θ)≤1 sin²(x-θ)≤1 [3y/√(y²+1)]²≤1 9y²/(y²+1)≤1 (8y²-1)/(y²+1)≤0 y²≤1/8 ...

如图所示、满意请采纳,谢谢。

解: y=(4-sinx)/(3-cosx) 3y-ycosx=4-sinx sinx-ycosx=4-3y √[1+(-y)²]sin(x-γ)=4-3y,(其中,tanγ=y) sin(x-γ)=(4-3y)/√(1+y²) -1≤sin(x-γ)≤1 -1≤(4-3y)/√(1+y²)≤1 (4-3y)²/(1+y²)≤1 (y -3/2)²≤3/8 (6-√6)/4≤y≤...

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