Problem 6 - Calculus for Biology and Medicine

Question

In Problems \(1-8\), find the gradient of each function. $$ f(x, y)=\tan \frac{x-y}{x+y} $$

Step-by-Step Solution

Answer

The gradient of the function is \( \left( \sec^2\left(\frac{x-y}{x+y}\right) \cdot \frac{2y}{(x+y)^2}, -\sec^2\left(\frac{x-y}{x+y}\right) \cdot \frac{2x}{(x+y)^2} \right) \).

Step 1: Calculate the Partial Derivative with Respect to x

For the function \( f(x, y) = \tan \frac{x-y}{x+y} \), consider \( u = \frac{x-y}{x+y} \) as a substitution. The derivative of \( \tan u \) with respect to \( u \) is \( \sec^2 u \). To find \( \frac{\partial f}{\partial x} \) using the chain rule, compute \( \frac{\partial u}{\partial x} \).\[ \frac{\partial u}{\partial x} = \frac{(x+y)(1) - (x-y)(1)}{(x+y)^2} = \frac{2y}{(x+y)^2} \] Thus, \[ \frac{\partial f}{\partial x} = \sec^2\left(\frac{x-y}{x+y}\right) \cdot \frac{2y}{(x+y)^2} \].

Step 2: Calculate the Partial Derivative with Respect to y

Again, using \( u = \frac{x-y}{x+y} \), find \( \frac{\partial u}{\partial y} \).\[ \frac{\partial u}{\partial y} = \frac{(x+y)(-1) - (x-y)(1)}{(x+y)^2} = -\frac{2x}{(x+y)^2} \] Therefore, the chain rule gives us \[ \frac{\partial f}{\partial y} = \sec^2\left(\frac{x-y}{x+y}\right) \cdot \left(-\frac{2x}{(x+y)^2}\right) \].

Step 3: Write the Gradient of the Function

The gradient of \( f(x, y) \) is given by the vector \( abla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}\right) \). Therefore, the gradient is:\[ abla f = \left( \sec^2\left(\frac{x-y}{x+y}\right) \cdot \frac{2y}{(x+y)^2}, -\sec^2\left(\frac{x-y}{x+y}\right) \cdot \frac{2x}{(x+y)^2} \right) \].

Key Concepts

Partial Derivatives

Partial derivatives involve differentiating a function with respect to one variable while keeping other variables constant. It’s like focusing on how one part of the situation changes at a time. In the case of the exercise at hand, we're looking at the function \( f(x, y) = \tan \frac{x-y}{x+y} \). The goal is to find the partial derivatives of this function with respect to \( x \) and \( y \).

• With respect to \( x \): We differentiate considering \( y \) as constant. The step-by-step breakdown includes substituting \( u = \frac{x-y}{x+y} \) to simplify our task. Understanding how the inside of the \( \tan \) function changes with \( x \) is vital, paving the way to find \( \frac{\partial f}{\partial x} \).
• With respect to \( y \): Similarly, we keep \( x \) constant. Substituting the same \( u \), find \( \frac{\partial f}{\partial y} \). Both of these derivatives together form the components of the gradient.

One of the key results here is that comprehending each variable's effect separately empowers you to build a clearer picture of the function's rate of change.

Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate compositions of functions. It enables us to find the derivative of functions like \( \tan u \) where \( u \) itself is a function of another variable. For the given exercise, substituting \( u = \frac{x-y}{x+y} \) highlights how important the chain rule becomes.

• We need to differentiate \( \tan u \). Its derivative is \( \sec^2 u \) with respect to \( u \). But since \( u \) is a composed function \( u(x, y) \), the chain rule requires us to multiply this by the partial derivative of \( u \) with respect to \( x \) or \( y \).

By understanding the chain rule, we connect the dots between differentiating a trigonometric function of a complex input and the simple components (\( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \)). This results in expressions for \( \frac{\partial f}{\partial x} \) and \( \frac{\partial f}{\partial y} \). Always keep in mind, when functions are nested within each other, the chain rule is your best approach.

Trigonometric Functions

Trigonometric functions like \( \tan \), \( \sin \), and \( \cos \) are pivotal in calculus, especially when dealing with periodic or circular phenomena. In the exercise, the function \( f(x, y) = \tan \frac{x-y}{x+y} \) is at the heart of trigonometry.

• The derivative of \( \tan u \) is \( \sec^2 u \). Recognizing this helps to correctly apply the chain rule. Beyond just the formula, knowing this derivative means predicting how tangents change, crucial for understanding waves and rotations.
• Trigonometric identities often simplify complex-looking expressions. Although not directly used in this exercise, they can transform problems dramatically when applied correctly.

Grasping their behavior benefits you across vast aspects of mathematics. Moreover, anytime you encounter trigonometric derivatives, remember they frequently tie back to real-world periodic events, making them familiar ground you stand on.