Answer :
[tex]f'(x)=\lim\limits_{h\to0}\frac{f(x-h)-f(x)}{h}\\\\f(x)=x^{-4}=\frac{1}{x^4}\\\\f'(x)=\lim\limits_{h\to0}\frac{(x+h)^{-4}-x^{-4}}{h}=\lim\limits_{h\to0}\frac{\frac{1}{(x+h)^4}-\frac{1}{x^4}}{h}=\lim\limits_{h\to0}\frac{\frac{x^4-(x+h)^4}{x^4(x+h)^4}}{h}\\\\=\lim\limits_{h\to0}\left[\frac{x^4-x^4-4x^3h-6x^2h^2-4xh^3-h^4}{x^4(x+h)^4}\cdot\frac{1}{h}\right]=\lim\limits_{h\to0}\frac{-4x^3h-6x^2h^2-4xh^3-h^4}{hx^4(x+h)^4}[/tex]
[tex]=\lim\limits_{h\to0}\frac{h(-4x^3-6x^2h-4xh^2-h^3)}{hx^4(x+h)^4}=\lim\limits_{h\to0}\frac{-4x^3-6x^2h-4xh^2-h^3}{x^4(x+h)^4}\\\\=\frac{-4x^3-6x^2\cdot0-4x\cdot0^2-0^3}{x^4(x+0)^4}=\frac{-4x^3}{x^4\cdot x^4}=\frac{-4}{x^5}\\\\\\D_f=D_{f'}=\mathbb{R}\ \backslash\ \{0\}[/tex]
[tex]=\lim\limits_{h\to0}\frac{h(-4x^3-6x^2h-4xh^2-h^3)}{hx^4(x+h)^4}=\lim\limits_{h\to0}\frac{-4x^3-6x^2h-4xh^2-h^3}{x^4(x+h)^4}\\\\=\frac{-4x^3-6x^2\cdot0-4x\cdot0^2-0^3}{x^4(x+0)^4}=\frac{-4x^3}{x^4\cdot x^4}=\frac{-4}{x^5}\\\\\\D_f=D_{f'}=\mathbb{R}\ \backslash\ \{0\}[/tex]
