Problem 38 - Calculus with Concepts in Calculus

Question

Find \(f_{x x}, f_{y y}\), and \(f_{z z}\) (where applicable). $$ f(x, y, z)=e^{x^{2}} \sin y z+\ln \left(x^{2}+y^{2}+z^{2}\right) $$

Step-by-Step Solution

Answer

\(f_{xx} = 2e^{x^2} \sin(yz) + 4x^2e^{x^2} \sin(yz) + \frac{2(y^2+z^2-x^2)}{(x^2+y^2+z^2)^2}\), \(f_{yy} = -e^{x^2} z^2\sin(yz) + \frac{2(x^2+z^2-y^2)}{(x^2+y^2+z^2)^2}\), \(f_{zz} = -e^{x^2} y^2\sin(yz) + \frac{2(x^2+y^2-z^2)}{(x^2+y^2+z^2)^2}\)."

Step 1: Identify the function components

The function given is \(f(x, y, z) = e^{x^2} \sin(yz) + \ln(x^2 + y^2 + z^2)\). It consists of two terms: \(e^{x^2} \sin(yz)\) and \(\ln(x^2 + y^2 + z^2)\). The first term depends on \(x\), \(y\), and \(z\), while the second term depends on \(x\), \(y\), and \(z\) as well.

Step 2: Compute the second partial derivative \(f_{xx}\)

First, find the first partial derivative with respect to \(x\): \(f_x = \frac{\partial}{\partial x}(e^{x^2} \sin(yz)) + \frac{\partial}{\partial x}(\ln(x^2 + y^2 + z^2))\). The derivatives are: - \(f_x = 2xe^{x^2} \sin(yz) + \frac{2x}{x^2 + y^2 + z^2}\).Next, find the second partial derivative with respect to \(x\):\(f_{xx} = \frac{\partial}{\partial x}(2xe^{x^2} \sin(yz)) + \frac{\partial}{\partial x}\left(\frac{2x}{x^2 + y^2 + z^2}\right)\). Process these separately:1. \(\frac{\partial}{\partial x}(2xe^{x^2} \sin(yz)) = 2e^{x^2} \sin(yz) + 4x^2e^{x^2} \sin(yz)\).2. To differentiate \(\frac{2x}{x^2 + y^2 + z^2}\), use the quotient rule, yielding \(\frac{2(y^2+z^2-x^2)}{(x^2+y^2+z^2)^2}\). So, the final expression for \(f_{xx}\) becomes:\(f_{xx} = 2e^{x^2} \sin(yz) + 4x^2e^{x^2} \sin(yz) + \frac{2(y^2+z^2-x^2)}{(x^2+y^2+z^2)^2}\).

Step 3: Compute the second partial derivative \(f_{yy}\)

First, find the first partial derivative with respect to \(y\): \(f_y = \frac{\partial}{\partial y}(e^{x^2} \sin(yz)) + \frac{\partial}{\partial y}(\ln(x^2 + y^2 + z^2))\).This becomes: - \(f_y = e^{x^2} z\cos(yz) + \frac{2y}{x^2 + y^2 + z^2}\).Next, find the second partial derivative with respect to \(y\): \(f_{yy} = \frac{\partial}{\partial y}(e^{x^2} z\cos(yz)) + \frac{\partial}{\partial y}\left(\frac{2y}{x^2 + y^2 + z^2}\right)\). 1. \(\frac{\partial}{\partial y}(e^{x^2} z\cos(yz)) = -e^{x^2} z^2\sin(yz)\).2. Differentiate \(\frac{2y}{x^2 + y^2 + z^2}\) leads to \(\frac{2(x^2+z^2-y^2)}{(x^2+y^2+z^2)^2}\).Thus, \(f_{yy} = -e^{x^2} z^2\sin(yz) + \frac{2(x^2+z^2-y^2)}{(x^2+y^2+z^2)^2}\).

Step 4: Compute the second partial derivative \(f_{zz}\)

First, find the first partial derivative with respect to \(z\): \(f_z = \frac{\partial}{\partial z}(e^{x^2} \sin(yz)) + \frac{\partial}{\partial z}(\ln(x^2 + y^2 + z^2))\).This becomes:- \(f_z = e^{x^2} y\cos(yz) + \frac{2z}{x^2 + y^2 + z^2}\).Next, find the second partial derivative with respect to \(z\): \(f_{zz} = \frac{\partial}{\partial z}(e^{x^2} y\cos(yz)) + \frac{\partial}{\partial z}\left(\frac{2z}{x^2 + y^2 + z^2}\right)\).1. \(\frac{\partial}{\partial z}(e^{x^2} y\cos(yz)) = -e^{x^2} y^2\sin(yz)\).2. Differentiating \(\frac{2z}{x^2 + y^2 + z^2}\) leads to \(\frac{2(x^2+y^2-z^2)}{(x^2+y^2+z^2)^2}\).So, \(f_{zz} = -e^{x^2} y^2\sin(yz) + \frac{2(x^2+y^2-z^2)}{(x^2+y^2+z^2)^2}\).

Key Concepts

Multivariable Calculus

In multivariable calculus, we explore functions that depend on multiple variables. Unlike single-variable functions, where we deal with one changing element, multivariable functions have several variables that can change simultaneously. This makes analyzing them a bit more complex but also fascinating. Consider

• The behavior of functions across a surface instead of along a line.
• The various dimensions affecting the function's output.

When you have a function like \[ f(x, y, z) = e^{x^2} \sin(yz) + \ln(x^2 + y^2 + z^2), \]components like \( e^{x^2} \sin(yz) \) and \( \ln(x^2 + y^2 + z^2) \) intertwine across axes, offering rich insights into geometrical and physical interpretations. With multivariable functions, you can:

• Model real-world phenomena with precision
• Gain insight into how systems react to changes in parameters

These learnings are widely applied in various fields, from engineering to computer graphics.

Second Partial Derivatives

The concept of second partial derivatives is vital in understanding the nature and behavior of multivariable functions. A partial derivative gives us the rate of change of a function with respect to one variable while keeping other variables constant.

The second partial derivative takes this a step further, offering how the rate of change itself changes. For a function \( f(x, y, z) \), calculating \( f_{xx} \), \( f_{yy} \), and \( f_{zz} \) involves:

• Finding the first partial derivative
• Taking the derivative of the result with respect to the same variable again

Why are these second-order derivatives helpful? They can tell us:

• Curvature, or how a surface bends in different directions
• Identify concave or convex regions

Practically, they help in optimizing functions and analyzing stability in systems.

Chain Rule in Calculus

The chain rule is an essential tool in calculus, especially when dealing with composite functions where one function is nested inside another. It allows us to differentiate complex expressions systematically. In the context of multivariable calculus, the chain rule adapts to handle situations where multiple variables are interdependent. With a function like \( f(x, y, z) = e^{x^2} \sin(yz) + \ln(x^2 + y^2 + z^2), \) the chain rule helps us navigate through the layers of dependence efficiently.

In essence, when you need to differentiate \( g(h(x)) \), you apply the rule as:

• Differentiate the outer function \( g \)
• Multiply by the derivative of the inner function \( h \)

This method shines in calculating derivatives of complex variables. Relating to partial derivatives, the insights from the chain rule clarify how independent changes in variables influence the function's outcome.