Problem 4 - Calculus for Biology and Medicine

Question

In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=x\left(x^{2}-y^{2}\right)^{2 / 3} $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f(x, y) = \left( (x^2 - y^2)^{2/3} + \frac{4x^2}{3(x^2 - y^2)^{1/3}}, -\frac{4xy}{3(x^2 - y^2)^{1/3}} \right) \).

Step 1: Understand the Problem

We are asked to find the gradient of the function \( f(x, y) = x(x^2 - y^2)^{2/3} \). The gradient of a scalar function \( f(x, y) \) is a vector consisting of its partial derivatives with respect to \( x \) and \( y \).

Step 2: Find the Partial Derivative with Respect to x

To find \( \frac{\partial f}{\partial x} \), apply the product rule to the function: \( f(x, y) = x \cdot (x^2 - y^2)^{2/3} \). Let \( u = x \) and \( v = (x^2 - y^2)^{2/3} \), so by the product rule, \( \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} v + u \frac{\partial v}{\partial x} \). Calculate each component:- \( \frac{\partial u}{\partial x} = 1 \).- \( \frac{\partial v}{\partial x} = \frac{2}{3}(x^2 - y^2)^{-1/3} \cdot 2x = \frac{4x}{3(x^2 - y^2)^{1/3}} \).Substitute back: \( \frac{\partial f}{\partial x} = (x^2 - y^2)^{2/3} + \frac{4x^2}{3(x^2 - y^2)^{1/3}} \).

Step 3: Find the Partial Derivative with Respect to y

To find \( \frac{\partial f}{\partial y} \), treat \( x \) as a constant and differentiate \( (x^2 - y^2)^{2/3} \) with respect to \( y \). Using the chain rule, this gives:- Let \( g = x^2 - y^2 \), then \( \frac{\partial g}{\partial y} = -2y \).- \( \frac{\partial f}{\partial y} = x \cdot \frac{2}{3}(x^2 - y^2)^{-1/3} \cdot (-2y) \).So, \( \frac{\partial f}{\partial y} = -\frac{4xy}{3(x^2 - y^2)^{1/3}} \).

Step 4: Combine Partial Derivatives to Form the Gradient

The gradient of the function is the vector consisting of the partial derivatives found in steps 2 and 3. Therefore, the gradient is:\[ abla f(x, y) = \left( (x^2 - y^2)^{2/3} + \frac{4x^2}{3(x^2 - y^2)^{1/3}}, -\frac{4xy}{3(x^2 - y^2)^{1/3}} \right) \].

Key Concepts

Partial Derivatives

Partial derivatives are derivatives of a function with respect to one variable while keeping other variables constant. They form the foundational concept behind gradients, which are critical in multivariable calculus. If you have a function like \( f(x, y) \), to find its partial derivatives, you differentiate \( f \) with respect to each variable separately. For example:

• For \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and differentiate \( f \) as if it's a function of \( x \) alone.
• For \( \frac{\partial f}{\partial y} \), treat \( x \) as constant and focus on taking the derivative with respect to \( y \).

Using partial derivatives, you can visualize how the function changes in different directions individually. It's akin to getting a road map where each route or path (variable) is considered one at a time. When working on the function \( f(x, y) = x(x^2 - y^2)^{2/3} \):

The gradient \( abla f \) is a vector that indicates the direction and rate of the steepest ascent from a point. This vector is made up of the partial derivatives \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Seeing each variable independently, yet considering their collective impact, helps uncover deeper insights into how functions behave, especially in fields like physics and engineering.

Product Rule

The product rule is a cornerstone of calculus when working with derivatives of products of two or more functions. If you have two functions, say \( u \) and \( v \), their product \( f = uv \) can be differentiated using the product rule:

The product rule formula is: \( \frac{d}{dx}(uv) = u'v + uv' \). This rule effectively tells you how to "share" the differentiation process between the two multiplying functions:

• Differentiate \( u \), leaving \( v \) untouched, and then vice versa.
• Add these two results together.

In our example function \( f(x, y) = x(x^2 - y^2)^{2/3} \), you can view it as being made up of \( u = x \) and \( v = (x^2 - y^2)^{2/3} \). Using the product rule helps in taking otherwise complex derivatives by breaking them into more manageable pieces. Such modular thinking simplifies not only math problems in academic settings but also real-world problems involving interconnected factors.

Chain Rule

Chain rule is the method used in calculus to differentiate composite functions. When a function is nested inside another function, the chain rule guides calculating these derivatives.

The chain rule formula generally is: if \( y = u(v(x)) \), then \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \). This insightfully allows steps to break down a complex differentiation problem into simpler parts:

• First, differentiate the outer function with respect to the inner function.
• Then, multiply by the derivative of the inner function with respect to x (or the variable of interest).

In the provided example function, while finding \( \frac{\partial f}{\partial y} \), realise \( g = x^2 - y^2 \) as an inner function within \( (x^2 - y^2)^{2/3} \). Application of the chain rule here handles derivatives of the nested structure. This approach is powerful in handling situations where one quantity depends on another and helps in effective problem resolution across fields like economics, biology, and more.

By practicing using the chain rule, you'll gain the ability to disentangle intricate chains of dependency, enabling meaningful analysis of how composite changes occur in multivariable functions.