Problem 52 - Calculus Early Transcendentals

Question

Find all the second partial derivatives. $$f(x, y)=\sin ^{2}(m x+n y)$$

Step-by-Step Solution

Answer

The second partial derivatives are \(2m^2 \cos(2(mx + ny))\), \(2n^2 \cos(2(mx + ny))\), and \(2mn \cos(2(mx + ny))\).

Step 1: Understand the Problem

We need to find all the second partial derivatives of the given function: \( f(x, y) = \sin^2(mx + ny) \). This involves calculating the first partial derivatives with respect to \(x\) and \(y\), and then finding the partial derivatives of the results.

Step 2: Find the First Partial Derivative with respect to x

The first step is to find the partial derivative of \(f(x, y)\) with respect to \(x\). Using the chain rule, we get:\[\frac{\partial f}{\partial x} = 2 \sin(mx + ny) \cdot \cos(mx + ny) \cdot m = m \sin(2(mx + ny)).\]

Step 3: Find the First Partial Derivative with respect to y

Next, we find the partial derivative of \(f(x, y)\) with respect to \(y\). Again using the chain rule, we have:\[\frac{\partial f}{\partial y} = 2 \sin(mx + ny) \cdot \cos(mx + ny) \cdot n = n \sin(2(mx + ny)).\]

Step 4: Find the Second Partial Derivative with respect to x

Now take the derivative of \(\frac{\partial f}{\partial x} = m \sin(2(mx + ny))\) with respect to \(x\). Apply the chain rule again:\[\frac{\partial^2 f}{\partial x^2} = m \cdot 2m \cos(2(mx + ny)) = 2m^2 \cos(2(mx + ny)).\]

Step 5: Find the Second Partial Derivative with respect to y

Take the derivative of \(\frac{\partial f}{\partial y} = n \sin(2(mx + ny))\) with respect to \(y\). This gives us:\[\frac{\partial^2 f}{\partial y^2} = n \cdot 2n \cos(2(mx + ny)) = 2n^2 \cos(2(mx + ny)).\]

Step 6: Find the Mixed Second Partial Derivative \(\frac{\partial^2 f}{\partial x \partial y}\)

Differentiate \(\frac{\partial f}{\partial x} = m \sin(2(mx + ny))\) with respect to \(y\):\[\frac{\partial^2 f}{\partial x \partial y} = m \cdot 2n \cos(2(mx + ny)) = 2mn \cos(2(mx + ny)).\]

Step 7: Find the Mixed Second Partial Derivative \(\frac{\partial^2 f}{\partial y \partial x}\)

Now differentiate \(\frac{\partial f}{\partial y} = n \sin(2(mx + ny))\) with respect to \(x\):\[\frac{\partial^2 f}{\partial y \partial x} = n \cdot 2m \cos(2(mx + ny)) = 2mn \cos(2(mx + ny)).\] Scince \(\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}\) by the symmetry of second partial derivatives, the calculations confirm this.

Key Concepts

Understanding First Partial Derivatives

The first partial derivatives capture the rate of change of a function with respect to each variable while keeping the other variables constant.
In the context of the function provided, we are looking to determine how changes in either \( x \) or \( y \) influence the function \( f(x, y) = \sin^2(mx + ny) \).

• To find the partial derivative with respect to \( x \), we treat \( y \) as constant and differentiate \( f(x, y) \) as if it were a single-variable function of \( x \).
• Similarly, for the partial derivative with respect to \( y \), treat \( x \) as constant.

In the solution, the first partial derivative with respect to \( x \) is calculated as \( \frac{\partial f}{\partial x} = m \sin(2(mx + ny)) \) using the chain rule. For \( y \), it is \( \frac{\partial f}{\partial y} = n \sin(2(mx + ny)) \).
These derivatives inform us about the slope of the function in the respective directions.

Chain Rule in Partial Derivatives

The chain rule is a fundamental principle used to differentiate composed functions, and it's especially useful in partial derivatives when dealing with functions of several variables.
When you have a composite function, like \( f(x, y) = \sin^2(mx + ny) \), the chain rule helps in finding the derivative by relating the derivative of the outer function to the derivative of the inner function.

• Here, \( u = mx + ny \) serves as the inner function, and the outer function is \( \sin^2(u) \).
• To apply the chain rule, first differentiate the outer function relative to the inner function.
• Then multiply it by the derivative of the inner function with respect to the variable of interest (either \( x \) or \( y \)).

Thus, the chain rule provides a structured way to manage the complexity of derivatives in multivariable calculus, making it indispensable for finding the partial derivatives in this exercise.

Mixed Partial Derivatives and Their Symmetry

Mixed partial derivatives involve differentiating a function first with respect to one variable and then with respect to another.
In the exercise, after obtaining the first partial derivatives of \( f(x, y) \), the next step is to find the mixed partial derivatives, like \( \frac{\partial^2 f}{\partial x \partial y} \) and \( \frac{\partial^2 f}{\partial y \partial x} \).

• These derivatives provide insight into what happens to the function when both variables, \( x \) and \( y \), change.
• The symmetry of mixed partial derivatives suggests that, under most conditions, \( \frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x} \).

This symmetry is known as Clairaut's theorem.
It holds when the function is well-behaved (i.e., it has continuous second partial derivatives), reassuring us about the interchangeability of the order of differentiation in a function like \( f(x, y) = \sin^2(mx + ny) \), as confirmed in the solution above.