Derivative Symmetric F13981

1. **State the problem:** We are given the function $f(x) = x^2 + x$ and asked to find the derivative using the limit definition: $$\lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}$$ 2. **Recall the formula:** This limit is a symmetric difference quotient, which also gives the derivative $f'(x)$: $$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x-h)}{2h}$$ 3. **Calculate $f(x+h)$ and $f(x-h)$:** $$f(x+h) = (x+h)^2 + (x+h) = x^2 + 2xh + h^2 + x + h$$ $$f(x-h) = (x-h)^2 + (x-h) = x^2 - 2xh + h^2 + x - h$$ 4. **Substitute into the limit expression:** $$\frac{f(x+h) - f(x-h)}{2h} = \frac{(x^2 + 2xh + h^2 + x + h) - (x^2 - 2xh + h^2 + x - h)}{2h}$$ 5. **Simplify the numerator:** $$= \frac{x^2 + 2xh + h^2 + x + h - x^2 + 2xh - h^2 - x + h}{2h}$$ $$= \frac{2xh + 2xh + h + h}{2h} = \frac{4xh + 2h}{2h}$$ 6. **Factor and cancel common terms:** $$= \frac{\cancel{2h}(2x + 1)}{\cancel{2h}} = 2x + 1$$ 7. **Take the limit as $h \to 0$:** Since the expression no longer depends on $h$, the limit is: $$\lim_{h \to 0} (2x + 1) = 2x + 1$$ **Final answer:** $$f'(x) = 2x + 1$$