5.已知f(x)二阶可导，且f(x)≠0，varphi(x)=lim_(tto0)((f(x+t))/(f(x)))^(1)/(sin t)，则varphi^prime(x)=__

令 $\varphi(x) = \lim_{t \to 0} \left( \frac{f(x+t)}{f(x)} \right)^{\frac{1}{\sin t}}$，取对数得
$\ln \varphi(x) = \lim_{t \to 0} \frac{\ln \left( \frac{f(x+t)}{f(x)} \right)}{\sin t} = \lim_{t \to 0} \frac{\ln(f(x+t)) - \ln(f(x))}{t} = \frac{f'(x)}{f(x)}.$
因此，$\varphi(x) = e^{\frac{f'(x)}{f(x)}}$。
对两边求导，利用链式法则和商法则，得
$\varphi'(x) = e^{\frac{f'(x)}{f(x)}} \cdot \frac{f''(x)f(x) - (f'(x))^2}{(f(x))^2} = \varphi(x) \cdot \frac{f''(x)f(x) - (f'(x))^2}{(f(x))^2}.$
答案：
$\boxed{\varphi(x) \frac{f''(x)f(x) - (f'(x))^2}{(f(x))^2}}$
（或等价表示：$\boxed{e^{\frac{f'(x)}{f(x)}} \frac{f''(x)f(x) - (f'(x))^2}{(f(x))^2}}$）