Question: Find the derivative of $y = x^2$ at $x = 3$ from first principles by:(a) sketching the parabola $y = x^2$(b) marking the points $R(3, 9)$ and $S(3 + h, (3 + h)^2)$ on it, where $h$ is some number.(c) finding the gradient $\frac{\Delta y}{\Delta x}$ of the line $RS$(d) evaluating the limit $\lim_{h \to 0} \frac{\Delta y}{\Delta x}$
1. **State the problem:**
Find the derivative of the function $$y = x^2$$ at the point $$x = 3$$ using first principles.
2. **Recall the definition of the derivative from first principles:**
The derivative of $$y = f(x)$$ at $$x = a$$ is given by the limit
$$
f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}
$$
This represents the slope of the tangent line at $$x = a$$.
3. **Mark the points on the parabola:**
- Point $$R$$ is at $$x=3$$, so $$R = (3, 3^2) = (3, 9)$$.
- Point $$S$$ is at $$x = 3 + h$$, so $$S = (3 + h, (3 + h)^2)$$.
4. **Find the gradient of the secant line $$RS$$:**
The gradient $$\frac{\Delta y}{\Delta x}$$ is
$$
\frac{(3 + h)^2 - 9}{(3 + h) - 3} = \frac{(3 + h)^2 - 9}{h}
$$
5. **Simplify the numerator:**
$$
(3 + h)^2 - 9 = (3 + h)(3 + h) - 9 = (9 + 6h + h^2) - 9 = 6h + h^2
$$
6. **Substitute back into the gradient expression:**
$$
\frac{6h + h^2}{h}
$$
7. **Cancel the common factor $$h$$:**
$$
\frac{\cancel{h}(6 + h)}{\cancel{h}} = 6 + h
$$
8. **Evaluate the limit as $$h \to 0$$:**
$$
\lim_{h \to 0} (6 + h) = 6
$$
9. **Interpretation:**
The derivative of $$y = x^2$$ at $$x = 3$$ is $$6$$, which means the slope of the tangent line to the parabola at $$x=3$$ is $$6$$.
---
**Final answer:**
$$
\boxed{6}
$$