Problem 48 - Calculus for Biology and Medicine

Question

Find the indicated partial derivatives. \(f(x, y)=\sin (3 x y) ; \frac{\partial^{2} f}{\partial y^{2}}\)

Step-by-Step Solution

Answer

\[ \frac{\partial^{2} f}{\partial y^{2}} = -9x^{2} \sin(3xy) \]

Step 1: Understand the Function

We start with the function given: \( f(x, y) = \sin(3xy) \). We need to find the second partial derivative of this function with respect to \( y \).

Step 2: First Partial Derivative with respect to y

To find \( \frac{\partial f}{\partial y} \), first differentiate \( f(x, y) \) with respect to \( y \), treating \( x \) as a constant. The derivative of \( \sin(u) \) where \( u = 3xy \) is \( \cos(u) \cdot \frac{du}{dy} \). Here, \( \frac{du}{dy} = 3x \). So, \( \frac{\partial f}{\partial y} = \cos(3xy) \cdot 3x = 3x \cos(3xy) \).

Step 3: Second Partial Derivative with respect to y

Now, differentiate \( \frac{\partial f}{\partial y} = 3x \cos(3xy) \) again with respect to \( y \). We apply the chain rule: the derivative of \( \cos(u) \) is \(-\sin(u) \cdot \frac{du}{dy} = -\sin(3xy) \cdot 3x \). Therefore, \( \frac{\partial^{2} f}{\partial y^{2}} = -9x^{2} \sin(3xy) \).

Key Concepts

Second Order Partial Derivative

In calculus, when we talk about derivatives of functions with more than one variable, we often encounter **partial derivatives**. These partial derivatives describe how a function changes as one specific variable changes, holding other variables constant. A **second order partial derivative** goes one step further. It's the derivative of a derivative. In simple terms, you're looking at how a partial derivative itself changes. This can help us understand the curvature or concavity of a surface described by a multivariable function.

Consider the function given: \[f(x, y) = \sin(3xy) \]We need the second partial derivative with respect to \( y \), which means we first find the derivative of \( f \) with respect to \( y \), and then again differentiate the result with respect to \( y \). This tells us about the behavior of the function's rate of change in the \( y \) direction.

To compute it:

• First derivative with respect to \( y \) gives us the rate of change of \( f \) concerning \( y \).
• Second derivative with respect to \( y \) analyzes how this rate is itself changing as \( y \) varies.

The final expression \(-9x^{2} \sin(3xy)\) describes this second-level change and gives deep insight into the behavior of the function overall.

Chain Rule

The **chain rule** is a vital concept in calculus for finding derivatives of composite functions. It helps to break complicated expressions into simpler parts that can be differentiated in steps. For a function composed of another function, such as \( f(x, y) = \sin(3xy) \), the chain rule is essential.When we differentiate \( \sin(3xy) \) with respect to \( y \), we're actually dealing with a composition of functions. Here, the outer function is \( \sin(u) \) and the inner one is \( u = 3xy \). We first differentiate the outer function \( \sin(u) \) with respect to its argument \( u \), giving us \( \cos(u) \), then multiply this by the derivative of the inner function \( u \) with respect to \( y \), which is \( 3x \).Thus, applying the chain rule helps us manage derivatives even in complex expressions, efficiently breaking them down into more straightforward parts. This approach not only simplifies calculations but also deepens understanding of how variables interact in a function.

Trigonometric Functions

In mathematics, **trigonometric functions** like sine and cosine are functions of an angle and are of profound importance in different areas like calculus, especially when dealing with oscillatory movements or waves. Here, we explore these in the context of calculus and derivatives.Given the function:\[f(x, y) = \sin(3xy) \]The sine function, \( \sin \), takes an argument \( 3xy \), where \( 3xy \) represents a scalar input to the sine function. Differentiating sine, a common task in calculus, involves using simple rules: the derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). This ability to transition between these functions when differentiating is fundamental whenever trigonometry enters differential calculus.Moreover, the later stage in our calculation requires applying these ideas to expressions like \(3xy\) inside our original function. Recognizing how trigonometric functions behave under differentiation is crucial. This knowledge allows us to simplify and solve complex calculus problems effectively, making trigonometric derivatives straightforward rather than initially intimidating.