If \( 1(x, y)=3 x e^{y}+x^{3} y^{2} \) then \( f_{y} \) is

To find \( f_{y} \), we need to compute the partial derivative of the function \( f(x,y) = 3x e^{y} + x^{3}y^{2} \) with respect to \( y \). Calculating \( f_{y} \): 1. Differentiate \( 3x e^{y} \) with respect to \( y \): \[ \frac{\partial}{\partial y}(3x e^{y}) = 3x e^{y} \] 2. Differentiate \( x^{3}y^{2} \) with respect to \( y \): \[ \frac{\partial}{\partial y}(x^{3}y^{2}) = x^{3}(2y) = 2x^{3}y \] Now, combine both results: \[ f_{y} = 3x e^{y} + 2x^{3}y \] Thus, \( f_{y} \) is: \[ f_{y} = 3x e^{y} + 2x^{3}y \] Now let’s dive into some additional information! The function \( f(x,y) = 3x e^{y} + x^{3}y^{2} \) can be viewed as a blend of exponential and polynomial terms, making it quite versatile in applications ranging from physics to economics. For instance, the term \( e^{y} \) can model growth phenomena, while \( y^{2} \) could represent varying returns in investment scenarios. The interplay between these terms creates fascinating dynamics! In solving problems involving partial derivatives, a common mistake is not carefully applying the product rule or ignoring terms that involve the variable of differentiation. Always remember to treat other variables as constants when calculating partial derivatives. Another frequent oversight is misplacing the exponential function’s base, so check your calculations double and triple times—precision is key in calculus!