y=tan根号x微分

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

两边对x求导：y'=(1+y')[sec(x+y)]^2得y'=[sec(x+y)]^2/{1-[sec(x+y)]^2}=1/{[cos(x+y)]^2-1}因此dy=dx/{[cos(x+y)]^

复合函数求导,tan的导数为sec,1/1+x^2的导数为-2x/(1+x^2)^2,综合有-sec^2(1/1+x^2)2x/(1+x^2)^2再问：1/1+x^2的导数？不是1+x^2的导数么再答

因为是偶函数,所以,sin(y+x)+根号3cos(x-y)=sin(y-x)+根号3cos(-x-y)所以,可以展开：sinycosx+cosysinx+根号3cosxcosy+根号3sinxsin

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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

令x+y=u,则dx+dy=du,代入换掉y,得du/dx=tanu+1,分离变量,得cosudu/(sinu+1)=dx,两边同时积分,得ln(sinu+1)=x+lnc所以通解为ln[sin(x+

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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偏z/偏x=1/2根号(1-x^2-y^2)×(-2x)偏z/偏y=1/2根号(1-x^2-y^2)×(-2y)所以dz=[1/2根号(1-x^2-y^2)×(-2x)]dx+[1/2根号(1-x^2

分解为y=5^u,u=lnv,v=tanx,用复合函数的求导法则；dy/dx=dy/du×du/dv×dv/dx用求导公式,dy/du=5^u×ln5,du/dv=1/v,dv/dx=(secx)^2

1y`=1+2xdy=(1+2x)dx2y`=2tanx(secx)^2dy=[2tanx(secx)^2]dx

zx=[√（x²+y²）-x²/√（x²+y²）]/（x²+y²）=y²/(x²+y²)^(3/2)

dz/dx=-3/2*(x^2+y^2)^(-3/2)*2x=-3x*(x^2+y^2)^(-3/2)dz/(dxdy)=-3x*(-3/2)*(x^2+y^2)^(-5/2)*2y=9xy*(x^2

y=arcsin√(1-x^2)y'=-x/(|x|√(1-x^2))∴dy=-xdx/(|x|√(1-x^2))当x>0dy=-dx/√(1-x^2)当x

设x/4=t则y=6tantt=x/4由复合函数求导公式：dy/dx=dy/dt*dt/dx=6sec^2(t)*(1/4)=3/2*sec^2(x/4)

y=x^2(cosx+√x),dy=[2x(cosx+√x)+x²(-sinx+1/2*1/√x)]dx=[2xcosx-x²sinx+2x√x+1/2*x√x]dx=[x(2co

y=1/x+√x=x^(-1)+x^(1/2)∴y'=(-1)*x^(-2)+(1/2)*x^(-1/2)=-1/x^2+1/(2√x)