Giúp giải bài đạo hàm này vs

a) C1: làm theo định nghĩa

a) y=f(x)=x³-3x²+2x+1

⇒ f'(x0)= lim((f(x)-f(x0))/(x-x0)) khi x→x0

⇔f'(x0)= lim(((x³-3x²+2x+1)-((x0)³-3(x0)²+2(x0)+1))/(x-x0)) khi x→x0

⇔f'(x0)=lim((x³-(x0)³-3x²+3(x0)²+2x-2(x0))/(x-x0)) khi x→x0

⇔f'(x0)=lim(((x-x0)(x²+x(x0)+(x0)²)-3(x-x0)(x+x0)+2(x-x0))/(x-x0)) khi x→x0

⇔f'(x0)=lim(x²+x(x0)+(x0)²-3(x+x0)+2) khi x→x0

⇔f'(x0)=(x0)²+(x0)(x0)+(x0)²-3(x0+x0)+2

⇔f'(x0)=3(x0)²-6(x0)+2

⇒y'=3x²-6x+2

C2: làm theo công thức

y'=(x³-3x²+2x+1)'=(x³)'+(-3x²)'+(2x)'+(1)'=3x²-3(x²)'+2(x)'=3x²-6x+2

b) C1: y'(x0)=f'(x0)=lim((f(x)-f(x0)/(x-x0)) khi x→x0

⇔y'(x0)=lim((-x³+3x+1+(x0)³-3(x0)-1)/(x-x0))khi x→x0

⇔y'(x0)=lim(-(x²+x(x0)+(x0)²)+3) khi x→x0

⇔y'(x0)=-3(x0)²+3

⇒y'=-3x²+3

C2: y=-x³+3x+1⇒y'=(-x³+3x+1)'=-3x²+3

c) C1:y'(x0)=lim (((f(x)-f(x0))/(x-x0)) khi x→x0

⇔y'(x0)=lim((((x^4)/4)-x²+1-(((x0)^4)/4)+x²-1))/(x-x0)) khi x→x0

⇔y'(x0)= lim (0.25*(x³+x²(x0)+x(x0)²+(x0)³)-(x+x0)) khi x→x0

⇔y'(x0)=(x0)³-2(x0)

⇒y'=x³-2x

C2:y=((x^4)/4)-x²+1⇒y'=(((x^4)/4)-x²+1)'=x³-2x

d)C1: y'(x0)=lim((-2x^4+1.5x²+1+2x^4-1.5x²-1)/(x-x0)) khi x→x0

⇔y'(x0)=lim(-2(x³+x²(x0)+x(x0)²+(x0)³)+1.5(x+x0)) khi x→x0

⇔y'(x0)=-8(x0)³+3(x0)

⇒y'=-8x³+3x

C2: y=-2x^4+1.5x²+1⇒y'=(-2x^4+1.5x²+1)'=-8x³+3x