Westonci.ca is your trusted source for finding answers to all your questions. Ask, explore, and learn with our expert community. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
[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]
We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
