Problem 6 - Calculus for Biology and Medicine

Question

In Problems 1-16, find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. $$ f(x, y)=\tan (x-2 y) $$

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = \sec^2(x - 2y) \), \( \frac{\partial f}{\partial y} = -2\sec^2(x - 2y) \).

Step 1: Understanding the Problem

We are asked to find the partial derivatives of the function \( f(x, y) = \tan(x - 2y) \) with respect to \( x \) and \( y \). This requires us to apply the rules of differentiation, treating one variable as a constant when differentiating with respect to the other.

Step 2: Differentiate with Respect to x

When differentiating \( f(x, y) = \tan(x - 2y) \) with respect to \( x \), treat \( y \) as a constant. The derivative of \( \tan(u) \) with respect to \( u \) is \( \sec^2(u) \). Thus, the derivative with respect to \( x \) is \( \frac{d}{dx}[\tan(x - 2y)] = \sec^2(x - 2y) \times 1 \) since the derivative of \( x \) with respect to \( x \) is 1.

Step 3: Differentiate with Respect to y

Now, differentiate \( f(x, y) = \tan(x - 2y) \) with respect to \( y \), treating \( x \) as a constant. The partial derivative is \( \frac{d}{dy}[\tan(x - 2y)] = \sec^2(x - 2y) \times (-2) \) because the derivative of \( -2y \) with respect to \( y \) is -2.

Step 4: Write Final Partial Derivatives

The partial derivative of \( f \) with respect to \( x \) is \( \frac{\partial f}{\partial x} = \sec^2(x - 2y) \) and with respect to \( y \) is \( \frac{\partial f}{\partial y} = -2\sec^2(x - 2y) \).

Key Concepts

Differentiation Rules

Differentiation is the process of finding a derivative. A derivative provides the rate of change of a function. Understanding differentiation rules is key to solving various calculus problems, including ones involving partial derivatives.

For partial derivatives, where we deal with functions of multiple variables, we apply similar rules as in basic differentiation but with some important considerations:

• Product Rule: Used when differentiating products of functions.
• Quotient Rule: Used for functions that are ratios of two simpler functions.
• Chain Rule: Essential in cases where composite functions are involved, helping to find the derivative of a function with respect to an inner function.

In the context of our original problem, which involves trigonometric functions, the chain rule plays a crucial role. The chain rule is applied when differentiating a nested trigonometric function, such as \( \tan(x-2y) \). Here the derivative \( \sec^2(u) \) is used, following the rule for \( \tan(u) \). Understanding when to apply each rule and breaking down the function methodically is critical.

Multivariable Calculus

Multivariable calculus extends the principles of single-variable calculus to functions with more than one variable. This branch of calculus is crucial for dealing with real-world problems where variables are interdependent.

Partial derivatives are the fundamental technique in multivariable calculus used to investigate the behavior of surfaces and curves in higher dimensions. When we find the partial derivative of a function \( f(x, y) \), we are interested in how the function changes as we vary one particular variable while keeping others constant.

In our example, we calculated partial derivatives of \( f(x, y) = \tan(x-2y) \) as follows:

• Partial Derivative with respect to \( x \): Treating \( y \) as constant gives \( \frac{\partial f}{\partial x} = \sec^2(x - 2y) \).
• Partial Derivative with respect to \( y \): Here, \( x \) is the constant, leading to \( \frac{\partial f}{\partial y} = -2\sec^2(x - 2y) \).

These calculations help us understand the behavior of \( f \) across different dimensions and provide a deeper insight into surface gradients and changes.

Trigonometric Functions

Trigonometric functions play a significant role in calculus, as they frequently appear in problems related to oscillatory motions and wave dynamics. Understanding their derivatives is essential for tackling such problems successfully.

For trigonometric functions like \( \tan(u) \), the derivative is \( \sec^2(u) \). This basic fact must be remembered along with other standard trigonometric derivatives:

• Derivative of \( \sin(u): \cos(u) \)
• Derivative of \( \cos(u): -\sin(u) \)
• Derivative of \( \tan(u): \sec^2(u) \)
• Derivative of \( \cot(u): -\csc^2(u) \)
• Derivative of \( \sec(u): \sec(u)\tan(u) \)
• Derivative of \( \csc(u): -\csc(u)\cot(u) \)

In our problem, the use of these derivatives, particularly \( \tan(u) \), allows us to effectively find the rates of change of our trigonometric expression in terms of \( x \) and \( y \). Mastery of these derivative rules enables a better understanding of how trigonometric expressions change and interact with each other within a broader calculus framework.