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Partial derivatives are a fundamental concept in calculus used to describe how a function changes as its input variables change independently. Imagine a surface or a curve. A partial derivative tells you how the height of the surface changes as you move in a particular direction (along one axis), while holding all other directions constant. For a function of two variables, such as \( e^x \ln(1+y) \), we compute two first-order partial derivatives: one with respect to \( x \) and one with respect to \( y \).When determining partial derivatives, you treat all variables except for the one you are differentiating with respect to as constants. This results in:

• The partial derivative with respect to \( x \), \( f_x = \frac{\partial f}{\partial x} = e^x \ln(1+y) \)
• The partial derivative with respect to \( y \), \( f_y = \frac{\partial f}{\partial y} = \frac{e^x}{1+y} \)

Both of these need to be evaluated at the point of approximation, which is the origin \((0,0)\) in this case. Evaluating gives \( f_x(0,0)=0 \) and \( f_y(0,0)=1 \). These derivatives help in forming Taylor expansions of the function.

A quadratic approximation of a function involves using terms up to the second degree to approximate the function around a point, such as the origin. Essentially, it includes a constant term, linear terms, and quadratic terms based on the function's partial derivatives.For our function \( f(x, y) = e^x \ln(1+y) \), the quadratic approximation relies on the second-order partial derivatives \( f_{xx}, f_{xy}, \) and \( f_{yy} \):

• \( f_{xx} = e^x \ln(1+y) \).
• \( f_{xy} = \frac{e^x}{1+y} \).
• \( f_{yy} = -\frac{e^x}{(1+y)^2} \).

Evaluating at the origin, we have values: \( f_{xx}(0,0)=0 \), \( f_{xy}(0,0)=1 \), and \( f_{yy}(0,0)=-1 \). Using these, the quadratic approximation of the function around the origin is:\[ f(x, y) \approx y + xy - \frac{1}{2}y^2 \]This approximation captures the local behavior of the function using only up to second-degree terms.

A cubic approximation is a step beyond quadratic approximation, which includes terms up to the third degree. It gives a more precise approximation of the function, especially when further away from the point of expansion.To achieve a cubic approximation for \( f(x, y) = e^x \ln(1+y) \), we consider third-order partial derivatives like \( f_{xxx}, f_{xxy}, f_{xyy}, \) and \( f_{yyy} \). Here's how some of them are defined and evaluated:

• \( f_{xxx} = e^x \ln(1+y) \)
• \( f_{xxy} = \frac{e^x}{1+y} \)
• \( f_{xyy} = -\frac{e^x}{(1+y)^2} \)
• \( f_{yyy} = \frac{2e^x}{(1+y)^3} \)

Evaluating at the origin gives \( f_{xxx}(0,0)=0 \), \( f_{xxy}(0,0)=1 \), \( f_{xyy}(0,0)=-1 \), and \( f_{yyy}(0,0)=2 \). Plugging these into the Taylor series formula for cubic approximation, we get:\[ f(x, y) \approx y + xy - \frac{1}{2}y^2 + \frac{1}{2}xy^2 + \frac{1}{3}y^3 \]This polynomial provides a finer and more accurate representation of the function near the origin, thanks to the additional cubic terms.