y'=2cosx+2x f‘（0）=2 y=2xf(x)=v(1+x) g(x)=x^2 gfx=v(x+1)^2=x+1x²-2x+1/x²-1=(x-1)^2/(x-1)(x+1)=（x-1）/（x+1)极限lim(x→1）x²-2x+1/x²-1=1-1/(1+1)=01/x(x-1)=1-1/x 原函数是x-lnx+c
曲线y=2sinx+x的2次方在（0,0）处的切线方程是?
y'=2cosx+2x f‘（0）=2 y=2xf(x)=v(1+x) g(x)=x^2 gfx=v(x+1)^2=x+1x²-2x+1/x²-1=(x-1)^2/(x-1)(x+1)=（x-1）/（x+1)极限lim(x→1）x²-2x+1/x²-1=1-1/(1+1)=01/x(x-1)=1-1/x 原函数是x-lnx+c