sin((pi 2)*x)的导数怎么求
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/16 21:07:28
![sin((pi 2)*x)的导数怎么求](/uploads/image/f/790947-27-7.jpg?t=sin%28%28pi+2%29%2Ax%29%E7%9A%84%E5%AF%BC%E6%95%B0%E6%80%8E%E4%B9%88%E6%B1%82)
[sin(x^5)]'=cos(x^5)×(x^5)'=cos(x^5)×5x^4=5x^4cos(x^5)
(x*cosx-sinx)/x^2
可以这么理解设y=x/2然后原题就变成求f(x)=siny的导数因为这里y是一个复合函数所以f(x)=y'(siny)'=1/2*cos(x/2)
f(x)=(sinx)^4f'(x)=4[(sinx)^3]*(sinx)'=4[(sinx)^3]*(cosx)
12cos(3x)cos(x)-4sin(3x)sin(x)
(sinx)^2求导是2sinxcosxsin(x^2)求导是2xcos(x^2)
∫sin²xdx=∫(1-cos2x)/2dx=1/2·∫1dx-1/2·∫cos2xdx=x/2-1/4·sin2x+C所以x/2-1/4·sin2x+C的导数是sin²x
y'=[-sinx(1-sinx)-cosx(-cosx)/(1-sinx)²=[-sinx+sin²x+cos²x]/(1-sinx)²=1/(1-sinx)
5sin^4x*cosx
y=sin^3x是复合函数可以设t=sinxt'=cosxy=t^3y'=3t^2*t'y'=3sin^2x*cosx
∵y=x^(sinx)∴lny=sinx*lnx两边求导:y'/y=cosx*lnx+(sinx)/x∴y'=y[cosx*lnx+(sinx)/x]=x^(sinx)[cosx*lnx+(sinx)
cos2x/(cosx+sinx)=(cos²x-sin²x)/(cosx+sinx)=(cosx+sinx)(cosx-sinx)/(cosx+sinx)=cosx-sinx求导
该题为复合函数求导,因此根据复合函数求导法则有:f(x)=u^2u=sinx所以f'(x)=(u^2)'u'=2uu'即f'(x)=(sin^2x)'=2sinxcosx=sin2x
(sin(x/2))'=(1/2)cos(x/2).再问:帮我做下这个好吗?
cosx谢谢o(︶︿︶)o再问:лл
[sin(1/x)]'=cos(1/x)(-1/x^2)=-1/x^2cos(1/x)
用链式法则:y=sin(πx)dy/dx=dsin(πx)/d(πx)*d(πx)/dx=cos(πx)*π(dx/dx)=cos(πx)*π=πcos(πx)
因为sin[2*(x/2)]=2sin(x/2)cos(x/2)所以x-sin(x/2)cos(x/2)=x-1/2sinx导数为1-1/2cosx
=[sin(1-x)]'=cos(1-x)*(1-x)'=-cos(1-x)
(x*sinx*cosx)'=(1/2xsin2x)'=1/2(sin2x+xcos2x*2)=1/2sin2x+xcos2x