Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { In Problems 1-10, find the gradient } \nabla f \text {. } $$ $$ f(x, y)=x e^{x y} $$

Step-by-Step Solution

Answer

\( \nabla f(x, y) = \left( e^{xy}(1 + xy), x^2 e^{xy} \right) \)

Step 1: Understand the Gradient

The gradient of a function is a vector that contains all of its partial derivatives. For a function \( f(x, y) \), the gradient is defined as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). We need to find the partial derivative of \( f(x, y) = x e^{xy} \) with respect to both \( x \) and \( y \).

Step 2: Find the Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), differentiate \( f(x, y) = x e^{xy} \) with respect to \( x \). This requires the product rule as \( f \) is a product of \( x \) and \( e^{xy} \). The derivative is \( \frac{d}{dx}[x] \cdot e^{xy} + x \cdot \frac{d}{dx}[e^{xy}] \). The derivative of \( x \) is 1, and the derivative of \( e^{xy} \) with respect to \( x \) (using the chain rule) is \( ye^{xy} \).Thus, \( \frac{\partial f}{\partial x} = e^{xy} + xye^{xy} = e^{xy}(1 + xy) \).

Step 3: Find the Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), differentiate \( f(x, y) = x e^{xy} \) with respect to \( y \). Here, treat \( x \) as a constant and differentiate \( e^{xy} \) directly.The derivative of \( e^{xy} \) with respect to \( y \) is \( xe^{xy} \), derived from the chain rule.Therefore, \( \frac{\partial f}{\partial y} = x^2 e^{xy} \).

Step 4: Write the Gradient Vector

Combine the partial derivatives to form the gradient vector. The gradient \( abla f \) is: \[ abla f(x, y) = \left( e^{xy}(1 + xy), x^2 e^{xy} \right) \].

Key Concepts

Partial Derivative

In the world of calculus, partial derivatives are a fascinating concept that focuses on how a function changes as we alter one of its variables at a time. Consider a function with multiple variables, such as \(f(x, y)\). The partial derivative of \(f\) with respect to \(x\) measures how \(f\) changes as \(x\) alone varies, keeping \(y\) constant. Similarly, the partial derivative with respect to \(y\) evaluates how the function behaves as \(y\) changes while \(x\) remains the same.

• For the function \(f(x, y) = x e^{xy}\), the partial derivative with respect to \(x\) is \(\frac{\partial f}{\partial x}\).
• The partial derivative with respect to \(y\) is \(\frac{\partial f}{\partial y}\).

By calculating partial derivatives, we gain insight into the function's sensitivity to changes in its variables. This is crucial for tasks such as optimization and modeling dynamic systems.

Product Rule

The product rule helps us differentiate functions that are products of two or more functions. If we have two functions, \(u(x)\) and \(v(x)\), that are multiplied together, the derivative of their product is found using the formula: \((uv)' = u'v + uv'\).
In our case, the function \(f(x, y) = x e^{xy}\) involves the variables \(x\) and \(e^{xy}\), where each behaves like a separate function. Applying the product rule here means:

• The derivative of \(x\) with respect to \(x\) is 1, while \(e^{xy}\) remains unaltered at first.
• Then, replace \(x\) with \(e^{xy}\), and find the partial derivative of the exponential component.

The product rule seamlessly combines with other rules to analyze even complex function behaviors.

Chain Rule

The chain rule is a powerful technique in calculus used to differentiate composite functions. It provides a method to compute the derivative of a function that is nested within another function. For example, given a function \(z = g(h(x))\), the chain rule states that the derivative of \(z\) with respect to \(x\) is \(g'(h(x)) \cdot h'(x)\).
In our function \(f(x, y) = x e^{xy}\), while using the product rule, we need to differentiate \(e^{xy}\).

• The function inside the exponent, \(xy\), requires differentiation, giving us \(y\) as its derivative with respect to \(x\).
• This derivative factor, \(y\), multiplies out with the derivative of \(e^{xy}\) itself.

Thus, seamlessly applying the chain rule allows us to tackle complex differentiations step by step.

Vector Calculus

Vector calculus extends the concepts of calculus to function in multiple dimensions. It is especially useful in physics and engineering and it beautifully describes things like the flow of fluids and electromagnetism. One of the main ideas is the gradient of a function, defined for scalar fields.

• The gradient, often represented as \(abla f\), is a vector emanating from a scalar function.
• For \(f(x, y) = x e^{xy}\), the gradient is \(abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right)\).

This vector points in the direction of the greatest rate of increase of the function and its magnitude is the rate of increase. Thus, understanding the gradient helps in optimizing functions and locating minima or maxima efficiently.