Problem 36 - Calculus Early Transcendentals: Pearson New International Edition

Question

If \(f(x, y)=\cos \left(2 x^{2}-y^{2}\right)\), find \(\partial^{3} f(x, y) / \partial y \partial x^{2} .\)

Step-by-Step Solution

Answer

The third mixed partial derivative is: \(4y (\cos(2x^2 - y^2) + 8x^2 \sin(2x^2 - y^2))\).

Step 1: Interpret the Exercise

The task is to find the third mixed partial derivative of the function \(f(x, y) = \cos(2x^2 - y^2)\). Specifically, we need to calculate \(\frac{\partial^3 f}{\partial y \partial x^2}\), which means differentiating twice with respect to \(x\) and once with respect to \(y\).

Step 2: Compute the First Partial Derivative with Respect to x

Find \(f_x(x, y)\), the first partial derivative of \(f\) with respect to \(x\). Let \(u = 2x^2 - y^2\), then \(f(x, y) = \cos(u)\). The chain rule gives:\[f_x(x, y) = -\sin(u) \cdot \frac{\partial u}{\partial x} = -\sin(2x^2 - y^2) \cdot 4x.\]

Step 3: Compute the Second Partial Derivative with Respect to x

Find \(f_{xx}(x, y)\), the second partial derivative.Use the product rule and chain rule on \(-4x \sin(2x^2 - y^2)\):\[f_{xx}(x, y) = -4 \sin(2x^2 - y^2) - 16x^2 \cos(2x^2 - y^2).\]

Step 4: Compute the Partial Derivative with Respect to y

Find \(f_{xxy}(x, y)\) by taking the partial derivative of \(f_{xx}(x, y)\) with respect to \(y\):Firstly, find\[\frac{\partial}{\partial y}(-4 \sin(2x^2 - y^2)) = 4y \cos(2x^2 - y^2).\]Secondly, find\[\frac{\partial}{\partial y}(-16x^2 \cos(2x^2 - y^2)) = 32x^2 y \sin(2x^2 - y^2).\]Therefore,\[f_{xxy}(x, y) = 4y \cos(2x^2 - y^2) + 32x^2 y \sin(2x^2 - y^2).\]

Step 5: Simplify Result

Simplify the obtained expression:\[f_{xxy}(x, y) = 4y \cos(2x^2 - y^2) + 32x^2 y \sin(2x^2 - y^2) = 4y \big(\cos(2x^2 - y^2) + 8x^2 \sin(2x^2 - y^2) \big).\]

Key Concepts

Chain Rule

The chain rule is a fundamental tool for differentiating composite functions. When faced with a complex function like \(f(x, y) = \cos(2x^2 - y^2)\), breaking it down into simpler parts makes differentiation easier to handle. In this example, you can think of \(u = 2x^2 - y^2\) as a simpler function. The chain rule lets you differentiate with respect to \(x\), even though \(x\) is not directly part of the cosine function.

Here's how it works: To find \(f_x(x, y)\), first identify the outer function (\(-\sin(u)\)) and the derivative of the inner function \(u\) with respect to \(x\), which in this case is \(4x\). So, you multiply the derivative of the outer function by the derivative of the inner function. This gives \(-\sin(2x^2 - y^2) \, \cdot \, 4x\).

• Identify the outer and inner functions properly.
• Remember that the chain rule helps differentiate by "chaining" the derivatives of these nested functions.
• Apply the derivative correctly by multiplying the derivative of the outer function with respect to the inner function.

Product Rule

The product rule is essential when differentiating products of two or more functions. In our exercise, after finding the first derivative with respect to \(x\), we apply the product rule to find \(f_{xx}(x, y)\). Let's dive deeper into what this means.

Suppose you have a function \(h(x) = g(x) \cdot k(x)\). To find its derivative, the product rule tells us\[h'(x) = g'(x)k(x) + g(x)k'(x).\]In the example, consider \(h(x, y) = -4x \sin(2x^2 - y^2)\). Applying the product rule:

• The derivative of \(-4x\) is \(-4\).
• The derivative of \(\sin(2x^2 - y^2)\) is found through the chain rule and comes out as \(-\cos(2x^2 - y^2)\cdot 4x\). Multiply these derivatives together as per the product rule.
• As a result, the derivative \(f_{xx}(x, y)\) combines both parts: \(-4 \sin(2x^2 - y^2) - 16x^2 \cos(2x^2 - y^2)\).

By following the product rule methodically, you can manage even complex derivatives.

Mixed Partial Derivatives

Mixed partial derivatives are particularly useful when you want to explore how a function changes with respect to multiple variables. In our function \(f(x, y) = \cos(2x^2 - y^2)\), the mixed derivative \(\frac{\partial^3 f}{\partial y \partial x^2}\) gives us a peek into how changes in both \(x\) and \(y\) simultaneously affect \(f(x, y)\).

To compute a mixed partial derivative, you differentiate with respect to one variable, and then again with others as indicated by the derivative's notation. Our goal here is to find \(f_{xxy}(x, y)\) by differentiating \(f_{xx}(x, y)\) with respect to \(y\).

• Start with the function obtained from differentiating twice with respect to \(x\).
• Apply the derivative with respect to \(y\) to each term separately. For terms like \(-4 \sin(2x^2 - y^2)\), apply the chain rule.
• For terms such as \(-16x^2 \cos(2x^2 - y^2)\), consider the chain rule again to find \(32x^2 y \sin(2x^2 - y^2)\).
• Simplify the result, understanding that each partial derivative provides insight into how the multi-variable function behaves.

By tackling each variable stepwise and logically, mixed partial derivatives let us see how multiple factors influence a function's overall change.