Problem 9 - Calculus and its applications

Question

Find \(f_{x}\) and \(f_{y}\). $$f(x, y)=e^{3 x-2 y}$$

Step-by-Step Solution

Answer

\(f_x = 3e^{3x-2y}\) and \(f_y = -2e^{3x-2y}\).

Step 1: Identify the partial derivatives

To find the partial derivatives of the function \(f(x, y)=e^{3x-2y}\), we need to differentiate the function separately with respect to \(x\) and \(y\).

Step 2: Differentiate with respect to x

To find \(f_x\), treat \(y\) as a constant and differentiate \(f(x, y)=e^{3x-2y}\) with respect to \(x\). We use the chain rule:\[\frac{d}{dx}[e^{3x-2y}] = e^{3x-2y} \cdot \frac{d}{dx}[3x-2y]\]Since \(2y\) is constant with respect to \(x\), \(\frac{d}{dx}[3x-2y] = 3\). Thus,\[f_x = 3 \cdot e^{3x-2y}\]

Step 3: Differentiate with respect to y

To find \(f_y\), treat \(x\) as a constant and differentiate \(f(x, y)=e^{3x-2y}\) with respect to \(y\). Again, we apply the chain rule:\[\frac{d}{dy}[e^{3x-2y}] = e^{3x-2y} \cdot \frac{d}{dy}[3x-2y]\]Since \(3x\) is constant with respect to \(y\), \(\frac{d}{dy}[3x-2y] = -2\). Thus,\[f_y = -2 \cdot e^{3x-2y}\]

Step 4: Summarize the results

The partial derivatives of the function are given by:- \(f_x = 3e^{3x-2y}\)- \(f_y = -2e^{3x-2y}\)

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus that helps us differentiate functions composed of other functions. This becomes very useful when handling functions with more complex expressions. For a function like \(f(x, y) = e^{3x - 2y}\), we see that the exponent \(3x - 2y\) is itself a function of \(x\) and \(y\). The chain rule allows us to differentiate by acknowledging this structure.
When using the chain rule, we identify the inner function and the outer function. The general idea is to:

• Differentiate the outer function while keeping the inner function intact.
• Then, multiply by the derivative of the inner function with respect to the variable of interest.

For partial derivatives, where we differentiate each variable separately, the chain rule aids in handling each variable independently. Understanding this rule is crucial when working with multi-variable calculus, as it ensures each variable is properly accounted for during differentiation.

Differentiation with Respect to x

Differentiation with respect to \(x\) involves treating all other variables as constants, focusing only on how the function changes along \(x\). For the function \(f(x, y) = e^{3x - 2y}\), this means considering \(y\) as a constant and differentiating with respect to \(x\).
Let's apply the chain rule here. First, recognize that the outer function is the exponential function \(e^{u}\), and the inner function is \(u = 3x - 2y\). Differentiating the outer function gives us \(e^{3x - 2y}\) because the derivative of \(e^u\) with respect to \(u\) is itself.
Next, multiply by the derivative of the inner function \(3x - 2y\) with respect to \(x\). Here, since \(y\) is constant, we calculate \(\frac{d}{dx}[3x - 2y] = 3\). This process yields:

• \(f_x = 3 \cdot e^{3x - 2y}\)

Understanding this differentiation step helps unravel the structure and influence of the variable \(x\) in multi-variable functions.

Differentiation with Respect to y

Differentiation of a function with respect to \(y\) focuses solely on the influence \(y\) has over the function, treating \(x\) as a constant. Looking back at our function \(f(x, y) = e^{3x - 2y}\), we make \(x\) a constant and differentiate with respect to \(y\).
Using the chain rule, we handle the exponential function with \(u = 3x - 2y\) as before. The derivative of \(e^{u}\) concerning \(u\) is \(e^{3x - 2y}\).
Then, the inner part \(3x - 2y\) is differentiated with respect to \(y\). Since \(3x\) is constant when differentiating with respect to \(y\), we calculate \(\frac{d}{dy}[3x - 2y] = -2\). This gives us:

• \(f_y = -2 \cdot e^{3x - 2y}\)

Grasping this differentiation method is vital for understanding how the variable \(y\) impacts such multi-variable functions.