Question

Given \(f(x, y)=\sin 4 x \cos 3 y\) find \(\frac{\partial^{2} f}{\partial x^{2}}, \frac{\partial^{2} f}{\partial y^{2}}\) and \(\frac{\partial^{2} f}{\partial x \partial y}\)

Short answer

Question: Calculate the second-order partial derivatives of the function \(f(x, y) = \sin 4x \cos 3y\). Answer: The second-order partial derivatives of the given function are: \(\frac{\partial^2 f}{\partial x^2} = -16\sin 4x\cos 3y\), \(\frac{\partial^2 f}{\partial y^2} = -9\sin 4x\cos 3y\), and \(\frac{\partial^2 f}{\partial x\partial y} = -12\cos 4x\sin 3y\).

Step-by-step solution

Find the first-order partial derivatives

First we need to find the first-order partial derivatives of the given function with respect to \(x\) and \(y\). The function \(f(x, y) = \sin 4x \cos 3y\) can be treated as the product of two functions, one only depending on \(x\) and the other only depending on \(y\). So let's compute these first-order partial derivatives. Using the chain rule for differentiation, we have $$\frac{\partial f}{\partial x} = \frac{d(\sin 4x)}{dx}\cos 3y = 4\cos 4x\cos 3y$$ and $$\frac{\partial f}{\partial y} = \sin 4x \frac{d(\cos 3y)}{dy} = -3\sin 4x\sin 3y$$

Find the second-order partial derivatives

Now that we have the first-order partial derivatives, we can proceed to compute the second-order partial derivatives. To find \(\frac{\partial^2 f}{\partial x^2}\), differentiate \(\frac{\partial f}{\partial x}\) with respect to \(x\): $$\frac{\partial^2 f}{\partial x^2} = \frac{d(4\cos 4x\cos 3y)}{dx} = -16\sin 4x\cos 3y$$ To find \(\frac{\partial^2 f}{\partial y^2}\), differentiate \(\frac{\partial f}{\partial y}\) with respect to \(y\): $$\frac{\partial^2 f}{\partial y^2} = \frac{d(-3\sin 4x\sin 3y)}{dy} = -9\sin 4x\cos 3y$$ To find \(\frac{\partial^2 f}{\partial x\partial y}\), differentiate \(\frac{\partial f}{\partial x}\) with respect to \(y\): $$\frac{\partial^2 f}{\partial x\partial y} = \frac{d(4\cos 4x\cos 3y)}{dy} = -12\cos 4x\sin 3y$$ So, we have obtained the second-order partial derivatives of the function as: $$\frac{\partial^2 f}{\partial x^2} = -16\sin 4x\cos 3y\ ,\ \frac{\partial^2 f}{\partial y^2} = -9\sin 4x\cos 3y\ ,\ \text{ and }\ \frac{\partial^2 f}{\partial x\partial y} = -12\cos 4x\sin 3y.$$

Key concepts

Chain Rule in Calculus

The chain rule is a fundamental technique in calculus used to differentiate composite functions. It's notably useful when you have a function inside another function. For functions of multiple variables, such as our example function \(f(x, y) = \sin 4x \cos 3y\), the chain rule allows us to differentiate with respect to one variable while treating others as constants.
In this exercise, we use the chain rule to find the partial derivatives with respect to \(x\) and \(y\).

• For \( \frac{\partial f}{\partial x} \), we differentiate \( \sin 4x \), applying the chain rule: \( \frac{d(\sin 4x)}{dx} = 4\cos 4x \).
• For \( \frac{\partial f}{\partial y} \), \( \cos 3y \) requires differentiation; the chain rule gives us \( \frac{d(\cos 3y)}{dy} = -3\sin 3y \).

Understanding the chain rule helps to manage complex expressions efficiently. It clarifies the sequence of differentiation when dealing with functions layered within other functions.

Second-Order Derivative

After finding first-order derivatives, calculating second-order derivatives requires us to differentiate again. This is vital to determine how the rate of change itself changes. This is not only crucial in theoretical calculus but also in fields like physics and engineering.
For the second-order partial derivatives:

• \( \frac{\partial^2 f}{\partial x^2} \) involves differentiating \( \frac{\partial f}{\partial x} \) again with respect to \(x\). We begin with \(4\cos 4x \cos 3y\) and apply the standard trigonometric differentiation method to get \(-16\sin 4x \cos 3y\).
• Similar logic applies for \( \frac{\partial^2 f}{\partial y^2} \), where \(-3\sin 4x \sin 3y\) becomes \(-9\sin 4x \cos 3y\) upon further differentiation with respect to \(y\).

The second-order cross partial derivative \( \frac{\partial^2 f}{\partial x \partial y} \) is an interesting case. It shows how the function's rate of change with one variable affects another; differentiating \(4\cos 4x \cos 3y\) with respect to \(y\) yields \(-12\cos 4x \sin 3y\). This demonstrates interaction between the variables.

Partial Derivatives

Partial derivatives are used when dealing with functions of multiple variables, as opposed to single-variable functions. They help us examine how changing one variable affects the function while keeping others constant.
In the function \(f(x,y) = \sin 4x \cos 3y\), we can separately compute the partial derivative with respect to each variable:

• \( \frac{\partial f}{\partial x} \) focuses on how \(f\) changes as \(x\) changes, given by \(4\cos 4x \cos 3y\).
• \( \frac{\partial f}{\partial y} \) reveals how \(f\) alters with changes in \(y\), resulting in \(-3\sin 4x\sin 3y\).

Partial derivatives provide insight into the structure and behavior of multivariable functions. Understanding how to take these derivatives is key in fields ranging from machine learning to thermodynamics, as it enables the study of complex dynamic systems.