令 y = a^x,所以 ln(y) = x ln(a) ==> (1/y) dy = ln(a) dx.因此,dy / dx = y ln(a) = (a^x) ln(a)
[SQR(3 - a^X)]' = 2( 3 - a^x) * (-a^x)' = -2 (3 - a^x) * (a^x) ln(a)
Therefore,f'(x) = [-2 (3 - a^x) * (a^x) ln(a)] / (a - 1).END.