1. **State the problem:** Find the derivative of the function $f(x) = |x + 7|$. 2. **Recall the formula:** The absolute value function $|u|$ can be written as $|u| = \sqrt{u^2}$, but more importantly, its derivative is given by: $$\frac{d}{dx} |u| = \frac{u}{|u|} \cdot \frac{du}{dx} \quad \text{for } u \neq 0$$ This means the derivative of $|x+7|$ depends on the sign of $x+7$. 3. **Apply the chain rule:** Let $u = x + 7$, then $f(x) = |u|$ and $\frac{du}{dx} = 1$. 4. **Write the derivative:** $$f'(x) = \frac{u}{|u|} \cdot 1 = \frac{x+7}{|x+7|}$$ 5. **Interpret the result:** - For $x + 7 > 0$, i.e., $x > -7$, $f'(x) = 1$. - For $x + 7 < 0$, i.e., $x < -7$, $f'(x) = -1$. - At $x = -7$, the derivative does not exist because the function has a sharp corner. **Final answer:** $$f'(x) = \begin{cases} 1 & \text{if } x > -7 \\ -1 & \text{if } x < -7 \end{cases}$$ or equivalently, $$f'(x) = \frac{x+7}{|x+7|} \quad \text{for } x \neq -7$$