Question

导数 设f(x)=a*e^x + b lnx,且F'(1) = e ,f'(-1) = 1/e ,求f'(0)的值.

Answer 1

F'(x)=a*e^x+b/x

F'(1)=a*e+b=e,f'(-1) = a/e-b=1/e

结合以上两式得出

(a-1)(e^2+1)=0

所以a=1,b=0

f(x)=e^x ,f'(x)=e^x

所以f'(0)=e^0=1