Problem 19 - 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 12.

Step 1: Find the partial derivative

Start by taking the partial derivative of the function \(f(x, y) = 3x^2ye^{x-y}\) with respect to \(x\). By utilizing the product rule and chain rule, \(f_{x}(x,y) = 3x^2y e^{x-y} + 2 \cdot 3xy \cdot e^{x-y} = 6x^2y e^{x-y} + 6xy e^{x-y}\)

Step 2: Evaluation at a point

Now, evaluate the derivative at the point (1,1). Plugging in 1 into \(f_{x}(x,y)\), we get \(f_{x}(1,1) = 6(1)^2(1)e^{1-1} + 6(1)(1)e^{1 - 1} = 6e^0 + 6e^0 = 6 + 6 = 12\)

Key Concepts

The Product Rule in Partial Derivatives

The product rule is essential when taking derivatives of products of functions. It is especially useful in partial derivatives of multivariable functions, like our function \( f(x, y)=3 x^{2} y e^{x-y} \). Here, the function is a product of two sub-functions: \( 3x^2y \) and \( e^{x-y} \). To apply the product rule, follow these steps:

• Differentiating the first sub-function, \( 3x^2y \), with respect to \( x \) yields \( 6xy \).
• The second function, \( e^{x-y} \), is treated as a constant while differentiating the first function. Simply multiply by the unevaluated second function to get the related term \( 6xye^{x-y} \).
• Next, differentiate the second function, \( e^{x-y} \), using the chain rule. Its derivative is \( e^{x-y} \) times \( 1 \) since \( \frac{d}{dx}(x-y) = 1 \).
• Multiply this result by the original first function \( 3x^2y \), yielding \( 3x^2y \, e^{x-y} \).

By combining these results, we use the product rule: \( f_{x}(x, y) = 6xy \, e^{x-y} + 3x^2y \, e^{x-y} \). This addition results from the derivatives of each sub-function plus the constants or other sub-functions as necessary. Then we simplify to get the final result, essential in evaluating derivatives.

Applying the Chain Rule

The chain rule is an indispensable tool in calculus, especially for computing derivatives of composite functions. It helps in determining the derivative of functions where variables are nested within other functions.Let's consider the application of the chain rule in our partial differentiation of \( f(x, y)=3 x^{2} y e^{x-y} \). This function involves an exponential term, \( e^{x-y} \), which contains \( x-y \) as a function within. To apply the chain rule here:

• Differentiate the outer function \( e^{u} \) with respect to \( u \), recognizing that it is simply \( e^{u} \), where \( u = x-y \).
• Determine the derivative of the inner function \( u = x-y \) with respect to \( x \), which is \( 1 \).
• Multiply these results to achieve the derivative of the composite function: \( e^{x-y} \times 1 = e^{x-y} \).

Thus, when the function unfolds through the chain rule, it simplifies the process of deriving the component linked to \( e^{x-y} \). This step is key in correctly applying the product rule to find \( f_{x} \). Every nested function needs a chain rule operation when differentiating with respect to the outer variable.

Thoroughly Evaluating Derivatives

Evaluating partial derivatives successfully without errors is an important skill in calculus. With the derivative expression \( f_{x}(x, y) = 6x^2y \, e^{x-y} + 6xy \, e^{x-y} \) derived,
we proceed to evaluation at specified points. For instance, the exercise required evaluating this expression at \((1,1)\).Step-by-step evaluation includes:

• Plug \( x = 1 \) and \( y = 1 \) into the derivative expression.
• Each term simplifies due to the nature of the exponential function; \( e^{1-1} = e^0 \) which equals \( 1 \).
• Calculate each expression separately: \( 6 \times 1^2 \times 1 \times 1 = 6 \) and \( 6 \times 1 \times 1 = 6 \).
• Sum these results yielding \( 12 \). This is the value of \( f_{x}(1,1) \).

The process requires careful substitution and simplification. This way, you're ensuring that each component is correctly evaluated after finding the derivative expression. Such practice builds confidence and ensures accuracy in solving calculus problems involving partial derivatives.