指數與對數之導函數
圖片參考：http://imgcld.yimg.com/8/n/AD01512151/o/151107090306413871933270.jpg問題32.33.38.44Ans:32.-1/x(x^2+1)?ln1033.1/xlnx38.2x?2^(x^2)?ln2+2^(2^x)?2^x?(ln2)^244.x^(x^x)?x^x[(1/lnx)^2+lnx+(1/x)]
(32) y =log10(√(x^2+1)/x) =log10[√(x^2+1)]-log10[x] y' =ln√(x^2+1)/ln10-lnx/ln10 =1/ln10*[ln√(x^2+1)-1nx] =1/ln10*[x/(x^2+1)-1/x] =-1/[x(x^2+1)ln10] (33) y =ln(log3√x) =ln[ln(√x)/ln3] =ln[ln(√x)]-ln(ln3) y' =1/ln(√x)*1/√x*1/2√x-0 =1/[2xln(√x)] [答案有出入](38) y' =[2^(x^2)*ln2*2x]+[2^(2^x)*ln2*(2^x)'] =[2^(x^2)*ln2*2x]+[2^(2^x)*ln2*[2^x*ln2]] =[2^(x^2)*ln2*2x]+[2^(2^x)*2^x*(ln2)^2] (44) (一) g=x^x lng=xlnx 1/g*g'=lnx+1 g'=(lnx+1)x^x (二) 如題 y=(x)^x^x lny=x^x*ln(x) 1/y*y'=(x^x)'ln(x)+x^xln'(x) 1/y*y'=[(lnx+1)x^x]*ln(x)+x^x*1/x y'={[(lnx+1)x^x]*ln(x)+x^x*1/x}(x)^x^x y'=(x)^x^x*x^x[(lnx)^2+lnx+1/x] [答案有出入]
其實你只要會公式就會算了==除了44題你要對整體取對數才有辦法把X弄下來
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