sin(y^2 x^2)
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/08 20:37:35
x=0:0.01:1;y=0;fori=1:20y=y+sin(i*x);endplot(y);
sin(x+y)sin(x-y)=-1/2(cos(x+y+x-y)—cos(x+y-x+y))=-1/2(cos2x—cos2y)=-1/2(1-2(sinx)^2-1+2(siny)^2)=(si
tan,正切;sin,正弦;cos,余弦tan(x+y)tan(x-y)=sin(x+y)/cos(x+y)*sin(x-y)/cos(x-y)=sin(x+y)sin(x-y)/[cos(x+y)c
左边=(sinxcosy+cosxsiny)(sinxcosy-cosxsiny)=sin²xcos²y-cos²xsin²y=sin²x(1-sin
dy/dx相当于对x进行求导:dy/dx=y'=2x*cos[sin(x^2)]*cos(x^2)由于:sinx=cosx,sin(x^2)=2x*cos(x^2)
你的意思是高次化低次吧利用公式cos2x=1-2sin(^2)X,sin(^2)X=(1-cos2x)/2sin(^3)(X/2)=sin(^2)(X/2)sin(X/2)=(1-cosx)/2*si
y'=(cos²x)'-(sin3^x)'=2cosx·(cosx)'-cos3^x·(3^x)'=2cosx·(-sinx)-cos3^x·(3^x·ln3)=-sin2x-ln3·cos
sin^2x+cos^2y=1/2∴sin^2x=1/2-cos^2y3sin^2x+sin^2y=3(1/2-cos^2y)+sin^2y=1.5-3cos^2y)+sin^2y又有sin^2y+c
sinx+siny+sinz-sin(x+y+z)=4sin[(x+y)/2]sin[(x+z)/2]sin[(y+z)/2]sinx+siny+sinz-sin(x+y+z)=2sin[(x+y)/
-2k=cos2x-cos2y=[2(cosx)^2-1]-[2(cosy)^2-1]=2[(cosx)^2-(cosy)^2]cos^2x-cos^2y=-k
(1)当y=C时,sin[(x+C)/2]=sin[(x-C)/2]移项,和差化积有2cos{[(x+C)/2+(x-C)/2]/2}sin{[(x+C)/2-(x-C)/2]/2}=0,即cos(x
对这样的隐函数求导数的时候,就把y看作x的函数,y对x求导就得到dy/dx所以原等式对x求导得到2xy²+x²*2y*dy/dx+siny+x*cosy*dy/dx=0于是化简得到
x=(x+y)/2+(x-y)/2y=(x+y)/2-(x-y)/2所以左边=cos[(x+y)/2+(x-y)/2]-cos[(x+y)/2-(x-y)/2]={cos[(x+y)/2]cos[(x
过程:先将括号里的当作一个整体,即求sinx的导数,所以是cos(2x+30度),再对括号里的求导,所以得2由复合函数的求导法则,知y=2cos(2x+30度)
y=sin²x+2sinxcosx+3cos²xy=(sin²x+cos²x)+2sinxcosx+(2cos²x-1)+1=1+sin2x+cos2
sin^2x+sin^2y-sin^2x*sin^2y+cos^2x*cos^2y=sin^2x-sin^2x*sin^2y+sin^2y+cos^2x*cos^2y=sin^2x*(1-sin^2y
我来给你解答,稍等再答:
令y=kx则limsin(y×x^2+y^4)/(x^2+y^2)=limsin[kx^3+(kx)^4]/[(1+k^2)*x^2]分子用等价无穷小替换=lim[k+(k^4)*x]*(x^3)/[