令f(x)=(2x-3)^5=a0+a1x+a2x^2+a3x^3+.+a5x^5
则f'(x)=10(2x-3)^4=a1+2a2x+3a3x^2+4a4x^3+5a5x^4
代入x=1,得：f'(1)=10=a1+2a2+3a3+4a4+5a5.