1. **State the problem:** Find the derivative of the function $f(x) = \frac{1}{x^2} + 1$ using first principles (the definition of the derivative). 2. **Recall the definition of the derivative:** $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ 3. **Substitute the function into the definition:** $$f'(x) = \lim_{h \to 0} \frac{\left(\frac{1}{(x+h)^2} + 1\right) - \left(\frac{1}{x^2} + 1\right)}{h} = \lim_{h \to 0} \frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h}$$ 4. **Simplify the numerator:** $$\frac{1}{(x+h)^2} - \frac{1}{x^2} = \frac{x^2 - (x+h)^2}{x^2 (x+h)^2}$$ 5. **Expand and simplify the numerator's numerator:** $$(x+h)^2 = x^2 + 2xh + h^2$$ So, $$x^2 - (x^2 + 2xh + h^2) = -2xh - h^2$$ 6. **Rewrite the difference quotient:** $$\frac{\frac{1}{(x+h)^2} - \frac{1}{x^2}}{h} = \frac{-2xh - h^2}{h \cdot x^2 (x+h)^2} = \frac{h(-2x - h)}{h \cdot x^2 (x+h)^2}$$ 7. **Cancel $h$ in numerator and denominator:** $$= \frac{-2x - h}{x^2 (x+h)^2}$$ 8. **Take the limit as $h \to 0$:** $$f'(x) = \lim_{h \to 0} \frac{-2x - h}{x^2 (x+h)^2} = \frac{-2x}{x^2 \cdot x^2} = \frac{-2x}{x^4} = -\frac{2}{x^3}$$ **Final answer:** $$f'(x) = -\frac{2}{x^3}$$