求y=根号x分之a b的微分
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设u=√(y/x)u'x=(-1/2)x^(-3/2)y^(1/2)u'y=(1/2)(xy)^(-1/2)那么原式变成了arctanu=(1/u^2)所以(u^2)arctanu=1两边取全微分得到
dy=e^(x^x)(e^(xlnx))'dx=e^(x^x)*(x^x)*(1+lnx)
x+y=-3x=-y-3xy=2(-y-3)y=2-y²-3y-2=0y²+3y+2=0(y+1)(y+2)=0y1=-1y2=-2一、当y1=-1时,x1=-2,根号y分之x加根
根号内必须大于等于0故有x-1≥0且1-x≥0即x≥1且x≤1所以x=1将x=1代回去得y=3然后将x,y代入所求式即可你的所求式表述不是很清楚,所以没办法帮你求了
dy=arcsinxdx+xdx/根号(1-x^2)+xdx/(根号1-x^2+e^2)
y=x/(1+√x)则y'=[x'*(1+√x)-x*[(1+√x)]']/(1+√x)²=[(1+√x)-x*1/2√x)/(1+√x)²=(2+√x)/(2+4√x+2x)dy
如果对x求导,则ln|x|=yln|y|,1/x=y'/y+yy'/y=y'/y+y',.对数求导法.如果对y求导,则ln|x|=yln|y|,x'/x=ln|y|+y/y,x'=y^y(1+ln|y
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+√(
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dz=(y+1/y)dx+(x-x/y^2)dy