✔ 最佳答案
(1) 指數函式的微分y=a^(nx) => ln(y)=nx*ln(a)兩邊微分: y'/y=n*ln(a)y'=n*ln(a)*y=n*ln(a)*a^(nx)
a=log(3) => y(x)=[a^(x^2-x)]^0.5=a^(x^2/2-x/2)ln(y)=(x^2/2-x/2)*ln(a) => y'/y=(x-1/2)*ln(a)y'=(x-1/2)*ln(a)*y=(x-1/2)*ln(a)*a^(x^2/2-x/2)=(x-1/2)*ln[log(3)]*[log(3)]^(x^2/2-x/2)
2014-05-13 04:50:22 補充:
(2) 指數函式的積分
由上題: y'=n*ln(a)*y=n*ln(a)*a^(nx)
y=∫n*ln(a)*a^(nx)*dx
=a^(nx)
=∫[a^(3x)+a^(-3x)]dx.....a=10
=∫a^(3x)*dx+∫a^(-3x)*dx
=[1/3*ln(a)]{∫3*ln(a)*a^(3x)*dx+∫3*ln(a)*a^(3x)*dx}
={a^(3x)+a^(-3x)}/[3*ln(a)]
={10^(3x)+10^(-3x)}/[3*ln(10)]
2014-05-13 16:41:12 補充:
x=0~1
y={[10^3+10^(-3)]-(1+1)}/{3*ln(10)}
=(1000+0.001-2)/{3*ln(10)}
=998.001/{3*ln(10)}
=332.667/ln(10)
=144.475