Problem 5 - APEX Calculus

Question

Evaluate \(f_{x}(x, y)\) and \(f_{y}(x, y)\) at the indicated point. $$ f(x, y)=x^{2} y-x+2 y+3 \text { at }(1,2) $$

Step-by-Step Solution

Answer

\(f_x(1, 2) = 3\) and \(f_y(1, 2) = 3\).

Step 1: Differentiate with respect to x

To find the partial derivative of \(f(x, y)\) with respect to \(x\), we treat \(y\) as a constant and differentiate \(f(x, y) = x^2y - x + 2y + 3\). The derivative is \(f_x(x, y) = \frac{\partial}{\partial x}(x^2 y) - \frac{\partial}{\partial x}(x) + \frac{\partial}{\partial x}(2y) + \frac{\partial}{\partial x}(3)\). Calculating each term, we get: \(\frac{\partial}{\partial x}(x^2 y) = 2xy\), \(\frac{\partial}{\partial x}(-x) = -1\), \(\frac{\partial}{\partial x}(2y) = 0\), \(\frac{\partial}{\partial x}(3) = 0\). Therefore, \( f_x(x, y) = 2xy - 1 \).

Step 2: Differentiate with respect to y

To find the partial derivative of \(f(x, y)\) with respect to \(y\), we treat \(x\) as a constant and differentiate. So, we have \(f_y(x, y) = \frac{\partial}{\partial y}(x^2 y) - \frac{\partial}{\partial y}(x) + \frac{\partial}{\partial y}(2y) + \frac{\partial}{\partial y}(3)\). Calculating each term, we get: \(\frac{\partial}{\partial y}(x^2 y) = x^2\), \(\frac{\partial}{\partial y}(-x) = 0\), \(\frac{\partial}{\partial y}(2y) = 2\), and \(\frac{\partial}{\partial y}(3) = 0\). Therefore, \( f_y(x, y) = x^2 + 2 \).

Step 3: Evaluate \(f_x(1, 2)\)

Substitute \(x = 1\) and \(y = 2\) into \(f_x(x, y) = 2xy - 1\): \(f_x(1, 2) = 2(1)(2) - 1 = 4 - 1 = 3\).

Step 4: Evaluate \(f_y(1, 2)\)

Substitute \(x = 1\) and \(y = 2\) into \(f_y(x, y) = x^2 + 2\): \(f_y(1, 2) = 1^2 + 2 = 1 + 2 = 3\).

Key Concepts

Partial Derivative with Respect to x

A partial derivative with respect to a particular variable indicates how a function changes as that variable increases while keeping the other variables constant. In our scenario, to find the partial derivative with respect to \(x\), we consider \(y\) as a constant. Let's break it down:

• For the term \(x^2y\), the derivative with respect to \(x\) is \(2xy\) because we treat \(y\) as constant, and the derivative of \(x^2\) is \(2x\).
• The term \(-x\) becomes \(-1\) since \(x\) differentiates to \(1\).
• Other terms, like \(2y\) and \(3\), become \(0\) because they don't involve \(x\).

Thus, the partial derivative \(f_x(x, y) = 2xy - 1\). This shows how the function \(f(x, y)\) changes solely with a change in \(x\).

Partial Derivative with Respect to y

The partial derivative with respect to \(y\) indicates how \(f(x, y)\) changes as \(y\) changes while keeping \(x\) constant. Now, let's differentiate \(f(x, y)\) with respect to \(y\):

• For \(x^2y\), treat \(x^2\) as a constant, making the derivative \(x^2\).
• Terms like \(-x\) differentiate to 0 since they don't involve \(y\).
• The term \(2y\) becomes 2, because the derivative of \(y\) is 1.
• The constant \(3\) becomes 0.

The calculation results in \(f_y(x, y) = x^2 + 2\).
This helps us understand how \(f(x, y)\) changes specifically with \(y\), independent of \(x\).

Evaluation at Specific Points

After computing the partial derivatives, the next step involves evaluating them at specified points to understand how the function behaves there. We have the point \((1,2)\):

• For \(f_x(1, 2)\), substitute \(x = 1\) and \(y = 2\) into \(2xy - 1\). The expression simplifies to \((2)(1)(2) - 1 = 3\). This value represents the rate of change of \(f(x,y)\) with respect to \(x\) at the point \((1,2)\).
• Similarly, calculate \(f_y(1, 2)\) by substituting into \(x^2 + 2\). Then, \(f_y(1, 2) = (1)^2 + 2 = 3\), showing the rate of change with respect to \(y\).

These evaluations help in understanding the local behavior and sensitivity of the function \(f(x, y)\) around point \((1,2)\).