Re: [微分] 求這題的推算過程
用的方法和你不同 但還是提供一下 ^^
f(x+y) = f(x)f(y)
d d
----f(x+y) = ----f(x)f(y)
dx dx
=> f'(x+y) = f'(x)f(y)
let x = 0
=> f'(0+y) = f'(0)f(y)
=> f'(y) = f(y)
=> f'(x) = f(x)
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