作业帮求微分dy.y=ln［ln(lnx)］
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/01 21:47:26
y'=(x^x)'+(ln(arctan5x)'设f(x)=x^xlnf(x)=xlnx1/f(x)f'(x)=lnx+1f'(x)=f(x)(lnx+1)=x^x(lnx+1)ln(arctan5x
等式两边同时求导得:2y*y'+y'/y=4*x^3-->y'=4y*x^3/(2y^2+1)y'=dy/dx-->dy=y'*dx=dx*4y*x^3/(2y^2+1)
y=lnu,其中u=sinxDY/DX=(dy/du)*(du/DX)=(1/u)cosx=cosx/sinx
dy=(1/secx乘secxtanx)dx=tanxdx
x/y=ln(y/x)x(-1/y^2)y'+1/y=x/y(-y/x^2+y'/x)(1/y+x/y^2)y'=1/y+1/x[(y+x)/y^2]y'=(x+y)/xyy'=y/x
F(x,y)=x+lny-y=0dF(x,y)=0=(∂F(x,y)dx/∂x)+(∂F(x,y)dy/∂y)dy/dx=-(∂F(x,y)
lnx=1/x这是公式,
y=e^(xlnx)+ln[arctan(5x)]dy/dx=e^(xlnx)[lnx+1]+1/arctan(5x)*[1+(5x)^2]^(-1)*5=x^x[lnx+1]+5/{arctan(5
这是复合函数的求导：y=√u,u=lnv,v=3x^2则y'=1/(2√u)*u'=1/(2√u)*1/v*v'=1/(2√u)*1/v*6x=1/(2√u)*1/(3x^2)*6x=1/(x√u)=
y=ln(tanx)/ln(sinx)dy/dx=[lnsinx.d/dx(lntanx)-lntanxd/dx(lnsinx)]/[ln(sinx)]^2=[lnsinx.(1/tanx)(secx
y=xe^(Cx+1),C为任意常数详细过程点下图查看
y=ln(sinx)y'=cosx/sinx=cotxy''=-1/sin²x∴y''=-1/sin²xdy=cotxdx
y=2ln(lnx)dy=y'dx=(2/lnx)*(1/x)dx=2/xlnxdx
y'＝1／（1＋x²）×反正切就是括号里的手机打不出来dy＝y'dx手机答好麻烦.给分.
y=ln(sinx)y'=cosx/sinx=cotxy''=-1/sin²x∴y''=-1/sin²xdy=cotxdx
ln²x=u,dy=dlnu=u'/u=2lnx*1/xdx/u=2/xlnxdx
y=[ln(1-x)^2]^2y'=2[ln(1-x)^2]*[ln(1-x)^2]'=2[ln(1-x)^2]*[2ln(1-x)]'=2[ln(1-x)^2]*2*1/(1-x)=4*[ln(1-
(e^-x)=-e^(-x)arcsinx^2=1/√(1-x^4)*(x²)'=2x/√(1-x^4)ln(sinx)=1/sinx*cosx=cotx所以dy=[-(e^-x)arcsi
dy=dx/(√(1+x^2))不好意思,我没办法将过程打出来