1. The problem is to express $\tan \alpha$ in terms of slopes $m$ and $m_1$ where $m_1 = \frac{dy}{dx}$. 2. The formula for the tangent of the angle between two lines with slopes $m$ and $m_1$ is given by: $$\tan \alpha = \frac{m_1 - m}{1 + m m_1}$$ 3. Here, $m_1$ represents the slope of one line (or derivative $\frac{dy}{dx}$), and $m$ is the slope of the other line. 4. This formula comes from the tangent subtraction formula in trigonometry and the definition of slope as $\tan \theta$. 5. To use this formula, substitute the known values of $m$ and $m_1$ into the expression and simplify. 6. For example, if $m_1 = \frac{dy}{dx}$ and $m$ is known, then: $$\tan \alpha = \frac{\frac{dy}{dx} - m}{1 + m \frac{dy}{dx}}$$ 7. This gives the tangent of the angle $\alpha$ between the two lines or curves at the point of interest. 8. Remember that this formula assumes the lines are not vertical (slopes are defined) and the denominator $1 + m m_1 \neq 0$ to avoid division by zero. Final answer: $$\boxed{\tan \alpha = \frac{m_1 - m}{1 + m m_1}}$$