Y=(X+1)^2=x^2+2x+1 对上式求导，x^2求导是2x，2x求导是2，1求导是0 所以对上式求导后结果为y'=2x+2 带入x=2，y'(x=2)=2*2+2=6
符号我不会打，这里假设lim代表h趋于0的极限： y'=lim [f(x+h)-f(x)]/h =lim [x+h+1/(x+h)-x-1/x]/h =lim [h-h/x(x+h)]/h =lim 1-1/(x^2+xh) =1-1/x^2
y=x^2+x 求导数 y‘=2x+1 x=2 y’=5 y=x(x+1)在x=2处的导数值为5
y=x-1/x与x轴的交点：(-1,0),(1,0) y'=1+1/(x^2),y'(-1)=y'(1)=2 所以所求切线方程：y=2(x+1)与y=2(x-1) 你的答案错了。 画个图给你看看：
看图````````````````````````````````
y=(x+1)^2*(x-1) =(x^2+2x+1)*(x-1)全部展开 =(x^3-x^2)+(2x^2-2x)+(x-1) 所以y'=3x^2-2x+4x-2x+x =3x^2+x 当x=1时,y'=3^2+1=10
just use quotient ,product rule and composition rule step by step the answer is straightforward.
1/x(x+1)的n阶导数，见附件：
设u=2x+2 f(x)=cosu f((x)=-sin(2x+2)*2 f((-1)=0
答案是4 解析如下 y'=2(x+1)(x-1)+(x+1)(x+1) =3*x*x+2*x-1 y'|x=1 =3+2-1=4 希望你能看明白
解:y=(x^2-x)(x^2+3x+2) =x^4+2x^3-x^2-2x 求导得 y'=4x^3+6x^2-2x-2 因此 y'(0)=-2
y=(x+1)/(x^2+2) =1/[(x+1) +3/(x+1) -2] x属于[1,3]时: (x+1) +3/(x+1)为增函数, 取值范围 =[7/2,19/4] 因此, 所求的值域为: [4/11, 2/3]
答案
y=e^(ylnx), 两边对x求导得 y'=e^(ylnx)*(y'lnx+y/x) =x^y*(y'lnx+y/x), ∴(1-x^y*lnx)y'=y*x^(y-1), ∴y'=[y*x^(y-1)]/(1-x^y*lnx).
y=(1+lnx)^x lny=xln(1+lnx) y'/y=ln(1+lnx)+x*[1/(1+lnx)]*(1/x)=ln(1+lnx)+1/(1+lnx) y'=[(1+lnx)^x][ln(1+lnx)+1/(1+lnx)]
