设$g(0)=g'(0)=0$，$$f(x)=\begin{cases} g(x)\sin{\frac{1}{x}} & x \neq 0, \\ 0 & x=0, \end{cases}$$求$f'(0)$.

$$f'(0)=\lim\limits_{x\to 0}\frac{g(x)\sin{\frac{1}{x}}-f(0)}{x-0}=\lim\limits_{x\to 0}\frac{g(x)\sin{\frac{1}{x}}}{x}=\lim\limits_{x\to 0}\frac{g(x)}{x}\lim\limits_{x\to 0}\sin{\frac{1}{x}}=g'(0)\lim\limits_{x\to 0}\sin{\frac{1}{x}}=0.$$

求$f(x)={{x}^x}^x$的导数。

记$f=f(x)$，两边同时取对数，有$\ln{y}=x^x\ln{x}$，记$u=x^x$，于是$\ln{u}=x\ln{x}$，两边同时求导$\frac{1}{u}\frac{du}{dx}=\ln{x}+1$,于是$$\frac{du}{dx}=x^x(\ln{x}+1)$$

对于$\ln{y}=x^x\ln{x}$两边同时对$x$求导，有$$\frac{1}{y}\frac{dy}{dx}=x^x(\ln{x}+1)\ln{x}+x^{x-1}$$

$$f'(x)=\frac{dy}{dx}=\left(x^x\ln{x}(\ln{x}+1)+x^{x-1}\right){{x}^x}^x=\left(\ln{x}(\ln{x}+1)+\frac{1}{x}\right)x^{\left(x^x+x\right)}.$$