求大佬写一下这个线代题目的步骤，蟹蟹啦 线代作业题：需要解答过程，不要直接给个答案，蟹蟹!
\u6c42\u5927\u4f6c\u5e2e\u5fd9\uff0c\u87f9\u87f9\u5566(\u8981\u8fc7\u7a0b)1.\u56e0\u4e3ax^4=x^0\u00d7x^4=x^2\u00d7x^2=x^3\u00d7x=x^4\u00d7x^0,
\u800c\u5728(2x-1)^4\u4e2d\u542bx^4,x^3,x^2,x,x^0\u7684\u9879\u7684\u7cfb\u6570\u5206\u522b\u4e3a:
C(0,4)2^4=2^4=16,C(1,4)2^3\u00d7(-1)=4\u00d72^3\u00d7(-1)=-2^5=-32,
C(2,4)2^2\u00d7(-1)^2=2^3=8,C(3,4)2\u00d7(-1)^3=-2^3=-8,C(4,4)\u00d7(-1)^4=1.
\u5728(2x+1)^6\u4e2d\u542bx^4,x^3,x^2,x,x^0\u7684\u9879\u7684\u7cfb\u6570\u5206\u522b\u4e3a:
C(2,6)2^4=15\u00d72^4=240,C(3,6)2^3=20\u00d72^3=160,C(4,6)2^2=15\u00d72^2=60,
C(5,6)2=6\u00d72=12,C(6,6)=1.
\u2234\u5c55\u5f00\u5f0f\u4e2d\u542bx^4\u7684\u7cfb\u6570\u4e3a:16\u00d71+(-32)\u00d712+8\u00d760+(-8)\u00d7160+1\u00d7240=-928
2.y'=2cos(3x+\u03c0/4)\u00d73=6cos(3x+\u03c0/4),\u2234y'(\u03c0/3)=6\u00d7cos(\u03c0+\u03c0/4)=-3\u221a2.
\u4e0a\u9762\u5199\u89e3\u91ca\u3002\u4e0b\u9762\u5728\u8ba1\u7b97\u3002\u5728\u7b54\u3002
详细过程如上所示
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