Problem 19 - College Algebra and Calculus: An Applied Approach

Question

Let \(f(x, y)=3 x^{2} y e^{x-y}\) and \(g(x, y)=\) \(3 x y^{2} e^{y-x}\). Find each of the following. $$ f_{x}(1,1) $$

Step-by-Step Solution

Answer

The value of \(f_{x}(1,1)\) is 9.

Step 1: Taking the derivative

The first step is to differentiate \(f(x,y)=3x^2y e^{x-y}\) partially with respect to 'x'. Remember the product and the chain rules should be applied here. Here, treat 'y' as a constant and focus on differentiating with respect to 'x'. Thus, the partial derivative \(f_x\) of this function is given by the sum of two terms: The derivative of the first term \(3x^2y\) times the second term \(e^{x-y}\), plus the derivative of the second term \(e^{x-y}\) times the first term \(3x^2y\). This gives \(f_{x} = 6xy e^{x-y} + 3x^2y e^{x-y}\).

Step 2: Substituting x=1 and y=1

The next step is to substitute \(x=1\) and \(y=1\) into the just obtained expression for \(f_{x}\): . Then, \(f_{x}(1,1) = 6(1)(1) e^{1-1} + 3(1)^2(1) e^{1-1} = 6*e^0 + 3*e^0 = 6 + 3 = 9 .\)

Key Concepts

Product Rule

The product rule is a crucial differentiation technique used when dealing with products of two functions. It's especially important when finding partial derivatives of multi-variable functions. When you have a function that is the product of two functions, like in our exercise where we have \(3x^2y\) and \(e^{x-y}\), the product rule helps find the derivative with respect to a variable like 'x'.

• To apply the product rule, you take the derivative of the first function while keeping the second one unchanged.
• Then, add it to the derivative of the second function while keeping the first function unchanged.

For example, when differentiating \(3x^2y \, e^{x-y}\) with respect to 'x', the product rule states:

• First derivative: \(\frac{d}{dx}[3x^2y] \, e^{x-y} = 6xy \, e^{x-y}\)
• Plus second derivative: \(3x^2y \frac{d}{dx}[e^{x-y}] = 3x^2y \, e^{x-y}\)

Therefore, combining these results using the product rule gives the partial derivative \(f_{x} = 6xy \, e^{x-y} + 3x^2y \, e^{x-y}\). This method ensures that all parts of a product are properly differentiated.

Chain Rule

The chain rule is essential when differentiating composite functions, such as an exponential function nested within another function. In our exercise, the term \(e^{x-y}\) needs the chain rule to differentiate effectively with respect to 'x'.

Here's how the chain rule simplifies the differentiation process:

• First, identify the "outer function." In \(e^{x-y}\), the outer function is the exponential \(e^u\) where \(u = x-y\).
• Next, identify the "inner function." Here, \(u = x-y\).
• Differentiate the outer function with respect to the inner function. This gives \(\frac{d}{du}[e^u] = e^u\).
• Differentiate the inner function with respect to 'x'. This gives \(\frac{d}{dx}[x-y] = 1\).
• Combine these using the chain rule, multiplying them together: \(e^{x-y} \cdot 1 = e^{x-y}\).

Essentially, the chain rule allows us to manage the complexity of differentiating deeply nested functions by breaking them into manageable parts. It ensures precise derivatives which are crucial for calculus fundamentals.

Exponential Function

Exponential functions, represented as \(e^x\) or \(e^{x-y}\) in our exercise, have unique differentiation rules that make them pivotal in calculus. The most fascinating aspect of exponential functions is how the derivative of an exponential function relates closely to its original form.

Key points about exponential functions in differentiation:

• The base, \(e\), is a constant approximately equal to 2.718, but its properties make it ideal for calculus.
• The derivative of \(e^x\) is unique because it remains \(e^x\); hence, differentiating exponentials doesn't alter their basic character.
• When dealing with \(e^{x-y}\) in our exercise, even after applying the chain rule, it simplifies back to the same exponential expression, demonstrating the robustness of exponential properties.

Understanding exponential functions and their manipulation via derivatives, like in the solution of the partial derivative \(f_x\), offers a peek into their critical role in science and mathematics, as they model growth and decay processes accurately.