用微积分的chainrule解(1+4x)^5(3+x-x^2)^8操难死了

Question

用微积分的chain rule解 (1+4x)^5(3+x-x^2)^8
操难死了

Answer 1

f(x)=(1+4x)^5(3+x-x^2)^8=(4x+1)^5(x^2-x-3)^8.令g(x)=4x+1,h(x)=x^2-x-3.令G(x)=x^5,H(x)=x^8.则f(x)=G(g(x))H(h(x)).则f'(x)=G'(g(x))g'(x)H(h(x))+G(g(x))H'(h(x))h'(x) =5(g(x))^4*4*(h(x))^8+(g(x))^5*8(h(x))^7*(2x-1) =20(1+4x)^4*(x^2-x-3)^8+8(1+4x)^5*(x^2-x-3)^7*(2x-1) =(1+4x)^4(x^2-x-3)^7[20(x^2-x-3)+8(1+4x)(2x-1)] =(1+4x)^4(x^2-x-3)^7[20(x^2-x-3)+8(8x^2-2x-1)] =4(1+4x)^4(x^2-x-3)^7(21x^2-9x-17)