1. **Problem:** Find the slope of the function $f(x) = 5x^3 - 4x^2 + 2x - 9$ using the definition of the derivative with the delta symbol and tangent lines. 2. **Formula:** The slope of the tangent line at a point $x$ is given by the derivative: $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ This represents the slope of the tangent line to the curve at $x$. 3. **Step 1: Calculate $f(x + \Delta x)$** $$f(x + \Delta x) = 5(x + \Delta x)^3 - 4(x + \Delta x)^2 + 2(x + \Delta x) - 9$$ Expand each term: $$5(x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3) - 4(x^2 + 2x\Delta x + (\Delta x)^2) + 2x + 2\Delta x - 9$$ 4. **Step 2: Simplify $f(x + \Delta x) - f(x)$** Subtract $f(x) = 5x^3 - 4x^2 + 2x - 9$: $$5x^3 + 15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 4x^2 - 8x\Delta x - 4(\Delta x)^2 + 2x + 2\Delta x - 9 - (5x^3 - 4x^2 + 2x - 9)$$ Cancel terms: $$15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8x\Delta x - 4(\Delta x)^2 + 2\Delta x$$ 5. **Step 3: Form the difference quotient and simplify:** $$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{15x^2\Delta x + 15x(\Delta x)^2 + 5(\Delta x)^3 - 8x\Delta x - 4(\Delta x)^2 + 2\Delta x}{\Delta x}$$ Cancel $\Delta x$: $$\frac{\cancel{\Delta x}(15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8x - 4\Delta x + 2)}{\cancel{\Delta x}} = 15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8x - 4\Delta x + 2$$ 6. **Step 4: Take the limit as $\Delta x \to 0$:** $$f'(x) = \lim_{\Delta x \to 0} (15x^2 + 15x\Delta x + 5(\Delta x)^2 - 8x - 4\Delta x + 2) = 15x^2 - 8x + 2$$ 7. **Final answer:** The slope of the tangent line to $f(x)$ at any point $x$ is: $$\boxed{f'(x) = 15x^2 - 8x + 2}$$ This slope tells us how steep the curve is at any $x$ value. --- Note: Since the user asked only for derivatives answers using delta symbol and tangent lines, only the first function's derivative is fully solved here as per instructions to solve only the first problem completely.