Problem 24 - Calculus for Biology and Medicine

Question

Find the gradient of each function. $$ f(x, y)=\cos \left(3 x^{2}-2 y^{2}\right) $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f(x, y) = \left( -6x \sin(3x^2 - 2y^2), 4y \sin(3x^2 - 2y^2) \right) \).

Step 1: Understanding the Gradient

To find the gradient of a function, we must compute the partial derivatives of the function with respect to each of its variables. The gradient of a function \( f(x, y) \) is given by \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). For our given function \( f(x, y) = \cos(3x^2 - 2y^2) \), we need to find \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \).

Step 2: Computing the Partial Derivative with Respect to x

First, calculate the partial derivative \( \frac{\partial f}{\partial x} \). Using the chain rule, we find \( \frac{\partial}{\partial x} \cos(3x^2 - 2y^2) = -\sin(3x^2 - 2y^2) \cdot \frac{\partial}{\partial x}(3x^2 - 2y^2) \). The derivative of the inner function with respect to \( x \) is \( 6x \). Hence, \( \frac{\partial f}{\partial x} = -6x \sin(3x^2 - 2y^2) \).

Step 3: Computing the Partial Derivative with Respect to y

Next, compute the partial derivative \( \frac{\partial f}{\partial y} \). Again using the chain rule, \( \frac{\partial}{\partial y} \cos(3x^2 - 2y^2) = -\sin(3x^2 - 2y^2) \cdot \frac{\partial}{\partial y}(3x^2 - 2y^2) \). The derivative of the inner function with respect to \( y \) is \( -4y \). Therefore, \( \frac{\partial f}{\partial y} = 4y \sin(3x^2 - 2y^2) \).

Step 4: Formulating the Gradient Vector

Now that we have both partial derivatives, we can form the gradient vector. The gradient of \( f(x, y) \) is \( abla f(x, y) = \left( -6x \sin(3x^2 - 2y^2), 4y \sin(3x^2 - 2y^2) \right) \).

Key Concepts

Gradient

The **gradient** is an important concept in multivariable calculus. It represents the vector of partial derivatives for a given function. Essentially, the gradient indicates the direction of the steepest increase of the function, and its magnitude tells us how steep this increase is. In the context of a two-variable function like ours, the gradient is a vector with two components:

• \( \frac{\partial f}{\partial x} \)
• \( \frac{\partial f}{\partial y} \)

For example, consider the function: \[ f(x, y) = \cos(3x^2 - 2y^2) \]To determine the gradient, we calculate the partial derivatives with respect to both \(x\) and \(y\). The resulting gradient vector \( abla f(x, y) \) then gives us critical information about the function's behavior. This can be summarized as \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculating the actual derivatives provides insight into how the function changes around any given point \((x, y)\).

Partial Derivatives

Understanding **partial derivatives** is crucial when dealing with functions of multiple variables. While a derivative in single-variable calculus gives the rate of change of a function with respect to one variable, partial derivatives extend this concept to functions with several variables.

Take our function, \( f(x, y) = \cos(3x^2 - 2y^2) \). We want to find the rate of change with respect to each variable independently.

• With respect to \(x\): The partial derivative \( \frac{\partial f}{\partial x} \) focuses on how \(f\) changes as \(x\) changes, keeping \(y\) constant. According to our solution, this derivative is \(-6x \sin(3x^2 - 2y^2)\).
• With respect to \(y\): Similarly, the partial derivative \( \frac{\partial f}{\partial y} \) describes changes in \(f\) as \(y\) varies, with \(x\) held steady. This yields \(4y \sin(3x^2 - 2y^2)\).

By calculating these derivatives, we can analyze how variations in individual variables affect the overall function.

Chain Rule

The **chain rule** is a fundamental tool for finding derivatives of composite functions. When we deal with multivariable functions, especially those that involve nested expressions, the chain rule helps us unpack and differentiate them.

For our function \( f(x, y) = \cos(3x^2 - 2y^2) \), notice the composite nature because of the cosine involving another function \((3x^2 - 2y^2)\). The chain rule indicates that to differentiate \( \cos(3x^2 - 2y^2) \) with respect to a variable, say \(x\), we perform two steps:

• Differentiate the outer function \( \cos(u) \) to get \(-\sin(u)\).
• Multiply by the derivative of the inner function \( 3x^2 - 2y^2 \), which gives \(6x\) for partial derivative with respect to \(x\).

Thus, the chain rule leads us to \[ \frac{\partial}{\partial x} \cos(3x^2 - 2y^2) = -6x \sin(3x^2 - 2y^2) \]This systematic approach can then be applied to find the partial derivative with respect to \(y\) similarly. The chain rule is a powerful technique that underpins much of the calculus used for analyzing complex, interdependent systems.