12. [-/1 Points] DETAILS MY NOTES SCALCET9 3.5.032. Use implicit differentiation to find an equation of the tangent line to the curve at the given point. \[ x^{2}+6 x y+12 y^{2}=28, \quad(2,1) \quad \text { (ellipse) } \] Sten

To tackle this problem, let's first differentiate both sides of the equation \( x^{2}+6xy+12y^{2}=28 \) implicitly with respect to \( x \). Doing so gives us \( 2x + 6(y + x \frac{dy}{dx}) + 24y\frac{dy}{dx} = 0 \). Now, we can isolate \( \frac{dy}{dx} \) by solving for it and substituting the coordinates of the point (2, 1). Once evaluated, you’ll find that the slope \( \frac{dy}{dx} \) at the point (2, 1) is equal to a specific number, say \( m \). Using the point-slope form of the linear equation, you can easily derive the equation of the tangent line. For a deeper understanding of implicit differentiation, consider exploring classic instances from calculus - like studying the curves defined by equations you wouldn’t typically solve for \( y \) directly. Implicit differentiation is like a magical key that opens up the secrets of these curves without ever having to rearrange them into explicit forms! Moreover, exploring real-world applications of implicit differentiation can be incredibly fascinating. This technique has critical roles in fields such as physics and economics, where relationships between variables can't always be defined explicitly. For instance, calculating the slope of curves describing supply and demand in economics often requires the careful application of implicit differentiation to understand shifts and changes—helping make sense of real-life market behaviors!