Problem 22 - Calculus for Biology and Medicine

Question

Find the gradient of each function. $$ f(x, y)=\tan \frac{x-y}{x+y} $$

Step-by-Step Solution

Answer

The gradient is \( \nabla f = \left( \sec^2 \frac{x-y}{x+y} \cdot \frac{2y}{(x+y)^2}, \sec^2 \frac{x-y}{x+y} \cdot \frac{-2x}{(x+y)^2} \right). \)

Step 1: Recall the Gradient Formula

The gradient of a function \( f(x, y) \) is a vector composed of its partial derivatives with respect to \( x \) and \( y \). It is represented as \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Differentiate with respect to \( x \)

To find \( \frac{\partial f}{\partial x} \), we need the derivative of \( \tan \frac{x-y}{x+y} \) with respect to \( x \). Using the chain rule and quotient rule, we have: \[ \frac{d}{dx} \left( \tan \frac{x-y}{x+y} \right) = \sec^2 \frac{x-y}{x+y} \cdot \frac{{(1)\cdot(x+y) - (x-y)\cdot(1)}}{(x+y)^2} = \sec^2 \frac{x-y}{x+y} \cdot \frac{2y}{(x+y)^2}. \] Thus, \( \frac{\partial f}{\partial x} = \sec^2 \frac{x-y}{x+y} \cdot \frac{2y}{(x+y)^2}. \)

Step 3: Differentiate with respect to \( y \)

Next, we find \( \frac{\partial f}{\partial y} \). Applying the chain rule and quotient rule: \[ \frac{d}{dy} \left( \tan \frac{x-y}{x+y} \right) = \sec^2 \frac{x-y}{x+y} \cdot \frac{-(1)\cdot(x+y) + (x-y)\cdot(1)}{(x+y)^2} = \sec^2 \frac{x-y}{x+y} \cdot \frac{-2x}{(x+y)^2}. \] Thus, \( \frac{\partial f}{\partial y} = \sec^2 \frac{x-y}{x+y} \cdot \frac{-2x}{(x+y)^2}. \)

Step 4: Compile the Gradient

The gradient of \( f(x, y) = \tan \frac{x-y}{x+y} \) is obtained by combining the partial derivatives found in the previous steps. Therefore, the gradient is: \[ abla f = \left( \sec^2 \frac{x-y}{x+y} \cdot \frac{2y}{(x+y)^2}, \sec^2 \frac{x-y}{x+y} \cdot \frac{-2x}{(x+y)^2} \right). \]

Key Concepts

Partial Derivatives

Partial derivatives are used to assess how a function changes with respect to one variable while keeping the other variables constant. This is crucial in multivariable calculus when you're dealing with functions that have more than one independent variable.
For a function like \( f(x, y) = \tan \frac{x-y}{x+y} \), finding the gradient involves calculating the partial derivatives with respect to both \( x \) and \( y \).
The process requires you to differentiate the function as if all other variables were constants.
Once you calculate the partial derivatives like \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \), you can form the gradient vector. This vector provides crucial information about the slope or steepness of the function at any given point.
Understanding and calculating partial derivatives are foundational skills for analyzing functions in multiple dimensions.

Chain Rule

When dealing with composite functions like \( \tan \frac{x-y}{x+y} \), the chain rule is essential for differentiation. This rule allows you to differentiate nested functions by working through them step-by-step.
Here's how it works: you first differentiate the outer function, keeping the inner function intact.
Then, you multiply this result by the derivative of the inner function.
This method helps you break down complex problems into manageable tasks.
In our example, the chain rule is applied to differentiate \( \tan \frac{x-y}{x+y} \). This involves calculating \( \sec^2 \left(\frac{x-y}{x+y}\right) \), the derivative of the tangent function, and then multiplying it by the derivative of the fraction \( \frac{x-y}{x+y} \).
Mastering the chain rule is key to advancing in calculus, as it frequently pops up in problems involving nested functions.

Quotient Rule

The quotient rule comes into play when differentiating functions that are ratios of two expressions. This rule is essential when you have a function \( f(x) = \frac{g(x)}{h(x)} \) and need to take its derivative.
The rule states that the derivative \( \frac{d}{dx}\left(\frac{g(x)}{h(x)}\right) \) is calculated as \( \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \).
This means you first differentiate the numerator and denominator separately.
Then, combine them according to the rule, ensuring the final operation results in division by the square of the denominator.
For our function \( \tan \frac{x-y}{x+y} \), the expression \( \frac{x-y}{x+y} \) involves the quotient rule. Here, the derivatives of \( x-y \) and \( x+y \) are crucial in simplifying and solving the problem.
Using the quotient rule is an efficient way to handle complicated fractions, making it an indispensable tool in calculus.