Problem 7 - Calculus for Biology and Medicine

Question

Find \(\partial f / \partial x\) and \(\partial f / \partial y\) for the given functions. \(f(x, y)=\cos ^{2}\left(x^{2}-2 y\right)\)

Step-by-Step Solution

Answer

\( \frac{\partial f}{\partial x} = -4x\cos(x^2 - 2y)\sin(x^2 - 2y) \), \( \frac{\partial f}{\partial y} = 4\cos(x^2 - 2y)\sin(x^2 - 2y) \).

Step 1: Identify the Inner Function

The given function is \( f(x, y) = \cos^2(x^2 - 2y) \). Here, the inner function is \( u = x^2 - 2y \). Our task is to differentiate \( f \) with respect to \( x \) and \( y \), which requires using the chain rule.

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

To find \( \frac{\partial f}{\partial x} \), we apply the chain rule. Start by differentiating the outer function \( \cos^2(u) \) with respect to \( u \) which gives \( 2\cos(u)(-\sin(u)) = -2\cos(u)\sin(u) \). Now differentiate the inner function \( u = x^2 - 2y \) with respect to \( x \), which gives \( \frac{\partial u}{\partial x} = 2x \). The partial derivative is then:\[ \frac{\partial f}{\partial x} = -2\cos(x^2 - 2y)\sin(x^2 - 2y) \cdot 2x = -4x\cos(x^2 - 2y)\sin(x^2 - 2y) \]

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

To find \( \frac{\partial f}{\partial y} \), we again use the chain rule. Differentiate the outer function \( \cos^2(u) \) with respect to \( u \) as before, giving \( -2\cos(u)\sin(u) \). Now differentiate the inner function \( u = x^2 - 2y \) with respect to \( y \), which gives \( \frac{\partial u}{\partial y} = -2 \). The partial derivative is then:\[ \frac{\partial f}{\partial y} = -2\cos(x^2 - 2y)\sin(x^2 - 2y) \cdot (-2) = 4\cos(x^2 - 2y)\sin(x^2 - 2y) \]

Key Concepts

Chain Rule

The chain rule is an essential tool when dealing with derivative problems involving compositions of functions. Particularly in multivariable calculus, this rule helps us differentiate complex functions by breaking them down into simpler parts. Consider the function \( f(x, y) = \cos^2(x^2 - 2y) \). Here, the cosine squared function is applied to an inner function \( u = x^2 - 2y \).

To differentiate \( f \) with respect to a variable, we apply the chain rule:

• Differentiating the outer function \( \cos^2(u) \) with respect to \( u \) yields \(-2\cos(u)\sin(u) \), as this uses the standard derivative of \( \cos(u) \).
• This result is then multiplied by the derivative of the inner function \( u \) with respect to the variable of interest.

In the given problem, this method was used to find both \( \partial f / \partial x \) and \( \partial f / \partial y \). By breaking the function into layers, each layer can be differentiated systematically, making problems more approachable.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of multiple variables. In the exercise at hand, the function \( f(x, y) \) depends on two variables, \( x \) and \( y \). This means we're dealing with a surface in three-dimensional space rather than a simple curve.

To analyze changes in \( f \), we look at partial derivatives. These represent how \( f \) changes as we vary one of the variables while keeping the other constant:

• \( \frac{\partial f}{\partial x} \) shows how \( f \) changes with small changes in \( x \), with \( y \) held constant.
• \( \frac{\partial f}{\partial y} \) captures the change in \( f \) with respect to \( y \), keeping \( x \) fixed.

By understanding these derivatives, we gain insight into the function's behavior in a multivariable context, crucial for fields such as physics and engineering.

Trigonometric Functions

Trigonometric functions like sine and cosine often appear in calculus problems, including those involving derivatives. Their periodic and oscillatory nature makes them significant in many applications that require modeling of waves and cycles.

For \( f(x, y) = \cos^2(x^2 - 2y) \), both the function and its derivatives involve trigonometric calculations:

• The derivative of \( \cos(u) \) with respect to \( u \) provides a basic tool for finding derivatives of composite trigonometric functions.
• In this exercise, we utilized \( -\sin(u) \), which is the derivative of \( \cos(u) \), to find the partial derivatives.

Understanding how to manipulate these trigonometric derivatives is crucial for solving problems effectively, particularly as functions get more complex.