91Ó°ÊÓ

Partial derivatives are a fundamental concept in multivariate calculus, crucial for understanding how a function changes with respect to each of its variables. Here, we dealt with the function \( f(x, y) = (y - x^2)(y - 3x^2) \). To find the partial derivative with respect to \( x \) (denoted \( f_x \)), we applied the product rule, which allows us to differentiate products of functions.

• The product rule states that for two functions \( u(x, y) \) and \( v(x, y) \), the derivative \( \frac{\partial}{\partial x}(uv) = u_x v + u v_x \).
• For \( f_x \), we differentiate both parts separately and get \( f_x = -2x(y-3x^2) - 3x(y-x^2) \).

Evaluating \( f_x \) at \((0,0)\) yields zero, confirming that the change in function value is initially zero in the \( x \) direction at the origin. Similarly, the partial derivative with respect to \( y \) (\( f_y \)) involves applying the product rule again:

• For the function \( f_y = (y - 3x^2) + (y - x^2) \), it simplifies by linear terms.

Evaluating at \((0,0)\) also results in zero, suggesting no immediate change in function value in the \( y \) direction. These derivatives are central to identifying and analyzing critical points.

A local minimum refers to a point where a function takes a smaller value than any other nearby points. To establish whether a point is a local minimum, we often examine behavior in different directions.In the given function \( f(x, mx) = m^2x^2 - 4mx^3 + 3x^4 \), replacing \( y \) with \( mx \) transforms it and allows us to evaluate the function along lines through the origin. This form reveals that the lowest degree term \( m^2x^2 \) is positive, confirming \( x=0 \) is a local minimum on every line \( y=mx \).However, when substituting \( y = 2x^2 \) to check variations along the parabola, the result was \( f(x, 2x^2) = -x^4 \). The resulting negatives as \( x \) approaches zero indicate that there's actually no local minimum at the origin with respect to the full function. This discrepancy shows that examining a function using only linear paths (lines) can be misleading.

An extremum is a general term that includes both minima and maxima of functions. To establish extremum points for a function like ours, we need to determine all directions of change and where they demonstrate no increase or decrease (stationary points).

• We firstly verified that at \((0,0)\), both \( f_x \) and \( f_y \) are zero, indicating a stationary point that could be an extremum.

Next, analyzing along paths such as lines and parabolas helps to identify the nature (minimization or maximization) at these critical points. Here, **line analysis suggested a minimum**, whereas the **parabola analysis contradicted it**. Thus, care must be taken to evaluate multiple paths and consider quadratic forms when confirming extrema.

The product rule for differentiation is pivotal in handling complex functions where variables are interdependent. This rule makes it possible to differentiate functions expressed as a product of two or more terms.For our function \( f(x, y) = (y - x^2)(y - 3x^2) \), using the product rule involved:

• Finding the derivative of each part independently. For example, in finding \( f_x \), one takes the derivative of each factor while treating the other as a constant.
• Combining these results to obtain expressions like \( -2x(y-3x^2) \) and \( -3x(y-x^2) \), cumulatively forming \( f_x \).

The rule simplifies complex computations by breaking them into smaller, manageable derivatives, showing how changes in one variable affect the whole function. It's integral to the analysis of multivariable calculus functions, making it indispensable for students delving into calculus.