第1854题:高阶微分
有 y=exlnxy=\mathrm{e}^x \ln xy=exlnx ,求 d5yd^5 yd5y .
A. d5y=d^5y=d5y= ex(lnx+5x−6x2\mathrm{e}^x \Big ( \ln x +\dfrac{5}{x} -\dfrac{6}{x^2}ex(lnx+x5−x26 +7x3−8x4)dx5+\dfrac{7}{x^3} -\dfrac{8}{x^4} \Big ) dx^5+x37−x48)dx5
B. d5y=d^5y=d5y= ex(lnx+5x−6x2\mathrm{e}^x \Big ( \ln x + \dfrac{5}{x} -\dfrac{6}{x^2} ex(lnx+x5−x26 +8x3−6x4)dx5+\dfrac{8}{x^3} -\dfrac{6}{x^4}\Big ) dx^5+x38−x46)dx5
C. d5y=d^5y=d5y= ex(lnx+5x+10x2+20x3\mathrm{e}^x \Big ( \ln x + \dfrac{5}{x} +\dfrac{10}{x^2} +\dfrac{20}{x^3} ex(lnx+x5+x210+x320 +30x4+24x5)dx5+\dfrac{30}{x^4} +\dfrac{24}{x^5} \Big ) dx^5+x430+x524)dx5
D. d5y=d^5y=d5y= ex(lnx+5x−10x2+20x3\mathrm{e}^x \Big ( \ln x + \dfrac{5}{x} -\dfrac{10}{x^2} +\dfrac{20}{x^3}ex(lnx+x5−x210+x320 −30x4+24x5)dx5-\dfrac{30}{x^4} +\dfrac{24}{x^5} \Big ) dx^5−x430+x524)dx5