微分y=ln²(1-x)-搜答案-搜狗做题网

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

Z=（1/2）ln(1+x²+y²)dz=(1/2)2x/(1+x²+y²)dx+(1/2)2y/(1+x²+y²)dy=x/(1+x&su

第一题,这是个隐函数,两边对x求导得：2y'-1=(1-y')*ln(x-y)+(x-y)*(1-y')/(x-y)=(1-y')*ln(x-y)+(1-y')所以[3+ln(x-y)]y'=ln(x

(1)y=3x^2-ln1/x=3x^2+lnxdy=6xdx+(1/x)dx=(6x+1/x)dx(2)y=e^(-x)cosxdy=-e^(-x)cosxdx-e^(-x)sinxdx=-e^(-

-((2x)/(1-x^2))dx;(-E^-x-Sin[3+x])dx;2Cos[2x]dx

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)

求导y=ln ln ln(x^2+1)

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)

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

dz=[-3ysin3xy+1/(1+x+y)]dx+[-3xsin3xy+1/(1+x+y)]dy

z偏x=-sin3xy*3y+1/(x+y+1)z偏y=-sin3xy*3x+1/(x+y+1)dz=[-sin3xy*3y+1/(x+y+1)]dx+[sin3xy*3x+1/(x+y+1)]dy

symsx>>y=log(x+sqrt(1+x^2));>>simple(diff(y)ans=1/(1+x^2)^(1/2)>>y=log(2*x+sqrt(1+x^2));>>simple(dif

y＝ln[x+√(1+x²)]∴y'＝[x+√(1+x²)]'/[x+√(1+x²)]＝[1+x/√(1+x²)]/[x+√(1+x²)]＝[x+√(

dy/dt=2t/(1+t²)dx/dt=1-[1/(1+t²)]=t²/(1+t²)dy/dx=(dy/dt)/(dx/dt)=2/t

解y=ln²(1-2x)y'=dy/dx=[ln²(1-2x)]'=2ln(1-2x)[ln(1-2x)]'(1-2x)'=2ln(1-2x)[1/(1-2x)(-2)=[-4ln

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-

dy=dx/(√(1+x^2))不好意思,我没办法将过程打出来

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