f(x)=(x²+3x+1)(x²+1)=?

Sol
f(x)=(x^2+3x+1)(x^2+1)
=x^2(x^2+1)+3x(x^2+1)+(x^2+1)
=(x^4+x^2)+(3x^3+3x)+(x^2+1)
=x^4+3x^3+2x^2+3x+1
f'(x)=4x^3+9x^2+4x+3
f’(1)=4+9+4+3=20
or
f(x)=(x^2+3x+1)(x^2+1)
f'(x)=(x^2+3x+1)*2x+(x^2+1)*(2x+3)
=(2x^3+6x^2+2x)+(2x^3+3x^2+2x+3)
=4x^3+9x^2+4x+3
f’(1)=20
or
f(x)=(x^2+3x+1)(x^2+1)
f’(1)=lim(x->1)_[f(x)－f(1)]/(x－1)
=lim(x->1)_[(x^2+3x+1)(x^2+1)-10]/(x－1)
=lim(x->1)_(x^4+3x^3+2x^2+3x+1－10)/(x－1)
=lim(x->1)_(x^4+3x^3+2x^2+3x－9)/(x－1)
１　３　２　３　－９｜
　　１　４　６　　９｜１
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f’(1)=lim(x->1)_(x^3+4x^2+6x+9)=20

f(x)=(x²+3x+1)(x²+1)
f(x)=x^4+3x^3+2x^2+3x+1
f'(x)=4x^3+6x^2+4x+3
f'(1)=4(1)^3+6(1)^2+4(1)+3
f'(1)=17

f(x)=(x²+3x+1)(x²+1)
f(x)=x^4 +3x^3+x^2+x^2+3x+1
f(x)=x^4 +3x^3+2x^2+3x+1

f'(x)=4x^3 +6x^2 +4x+3
f'(1)=4(1)^3 +6(1)^2+4(1)+3
f'(1)=17