91Ó°ÊÓ

Partial derivatives are a fundamental concept in calculus, especially when dealing with functions of several variables like our function, \(f(x, y) = \sin x \cos y\). Unlike ordinary derivatives, which deal with single-variable functions, partial derivatives measure the rate of change of a multivariable function with respect to one variable while keeping the others constant.
When computing partial derivatives, it's essential to treat other variables as constants. For instance, when finding \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant. Conversely, when finding \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant.
In the solution, we computed:

• \( \frac{\partial f}{\partial x} = \cos x \cos y \)
• \( \frac{\partial f}{\partial y} = -\sin x \sin y \)

This process is crucial for deriving more complex expansions like the Taylor series.

The gradient is a vector that contains all of a function's first-order partial derivatives. It points in the direction of the greatest rate of increase of the function. For a two-variable function \(f(x, y)\), the gradient is represented as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).
In our example, the gradient at the origin (0,0) is given by evaluating the partial derivatives at this point:

• \( \frac{\partial f}{\partial x}(0, 0) = 1 \)
• \( \frac{\partial f}{\partial y}(0, 0) = 0 \)

Thus, \( abla f(0, 0) = (1, 0) \).
The gradient helps in formulating the linear part of the Taylor expansion. It's like the slope of a function when extended to multiple dimensions, guiding us in constructing approximations.

The Hessian matrix is a square matrix of second-order partial derivatives of a function. It provides additional information about the curvature and concavity of a multivariable function. For our function \(f(x, y)\), the Hessian \(H_f(x, y)\) is:\[H_f = \begin{bmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \ \frac{\partial^2 f}{\partial y \partial x} & \frac{\partial^2 f}{\partial y^2}\end{bmatrix}\]
From our calculations in the example:

• \( \frac{\partial^2 f}{\partial x^2}(0, 0) = 0 \)
• \( \frac{\partial^2 f}{\partial y^2}(0, 0) = 0 \)
• \( \frac{\partial^2 f}{\partial x \partial y}(0, 0) = 0 \)

At the origin, this Hessian simplifies to a matrix of zeros, indicating a lack of curvature, contributing nothing more to the quadratic approximation.

Quadratic approximation is a way to approximate a multivariable function using terms up to the second order. The Taylor series formula incorporates up to the second derivatives to form the quadratic approximation.
For our example, the quadratic approximation of \(f(x, y)\) around the origin is derived using the formula:\[f(x, y) \approx f(0,0) + \frac{\partial f}{\partial x}(0,0) x + \frac{\partial f}{\partial y}(0,0) y + \frac{1}{2}\left( \frac{\partial^2 f}{\partial x^2}(0,0) x^2 + 2\frac{\partial^2 f}{\partial x \partial y}(0,0) xy + \frac{\partial^2 f}{\partial y^2}(0,0) y^2 \right)\]
Plugging in the calculated values, the result is simply:\[f(x, y) \approx x\]
This indicates that near the origin, the function \(f(x, y)\) behaves almost like the function of \(x\), neglecting any contribution from \(y\) or cross-terms.

Cubic approximation extends the idea of the Taylor series to include terms up to third order. These terms involve third derivatives, giving a more precise approximation than the quadratic version.
In our example, we use the third derivatives to form the cubic approximation:\[f(x, y) \approx x + \frac{1}{3!} \left(\frac{\partial^3 f}{\partial x^3}(0,0)x^3 + 3 \frac{\partial^3 f}{\partial x^2 \partial y}(0,0)x^2 y + 3 \frac{\partial^3 f}{\partial x \partial y^2}(0,0)x y^2 + \frac{\partial^3 f}{\partial y^3}(0,0)y^3 \right)\]
The computed third derivatives give us:

• \(\frac{\partial^3 f}{\partial x^3}(0,0) = -1\)
• \(\frac{\partial^3 f}{\partial x^2 \partial y}(0,0) = 0\)
• \(\frac{\partial^3 f}{\partial x \partial y^2}(0,0) = 1\)
• \(\frac{\partial^3 f}{\partial y^3}(0,0) = 0\)

Plugging these into the formula, we derive the cubic approximation:\[f(x, y) \approx x - \frac{1}{6}x^3 + \frac{1}{2}xy^2\]
This approximation includes higher-order terms, providing a finer representation of \(f(x, y)\) near the origin.