https://file.aiquickdraw.com/mathbot/file/1729626300870avt5bfum.jpg

We are given two functions \( f(s,t) = (s - t + 2, e^{2s + 3t}, s \cos t) \) and \( g(x, y, z) = (xyz, x^2 + z^3) \), and we are tasked with finding the derivative \( D(f \circ g)(0, 0, 0) \) using the chain rule.

### Step 1: Understanding the Composition of Functions

We need to compute the derivative of the composition \( f(g(x, y, z)) \). The chain rule for multivariable functions states that:

\[
D(f \circ g)(x, y, z) = Df(g(x, y, z)) \cdot Dg(x, y, z)
\]

Where \( Df \) is the Jacobian matrix of \( f \), and \( Dg \) is the Jacobian matrix of \( g \).

### Step 2: Find \( Dg(x, y, z) \)

The Jacobian matrix \( Dg(x, y, z) \) is given by the partial derivatives of \( g \):

\[
g(x, y, z) = (xyz, x^2 + z^3)
\]

So the partial derivatives are:

\[
Dg(x, y, z) =
\begin{pmatrix}
\frac{\partial}{\partial x}(xyz) & \frac{\partial}{\partial y}(xyz) & \frac{\partial}{\partial z}(xyz) \\
\frac{\partial}{\partial x}(x^2 + z^3) & \frac{\partial}{\partial y}(x^2 + z^3) & \frac{\partial}{\partial z}(x^2 + z^3)
\end{pmatrix}
=
\begin{pmatrix}
yz & xz & xy \\
2x & 0 & 3z^2
\end{pmatrix}
\]

At the point \( (0, 0, 0) \), this becomes:

\[
Dg(0, 0, 0) =
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]

### Step 3: Find \( Df(s, t) \)

Next, we need the Jacobian matrix of \( f(s, t) = (s - t + 2, e^{2s + 3t}, s \cos t) \). The partial derivatives of \( f \) are:

\[
Df(s, t) =
\begin{pmatrix}
\frac{\partial}{\partial s}(s - t + 2) & \frac{\partial}{\partial t}(s - t + 2) \\
\frac{\partial}{\partial s}(e^{2s + 3t}) & \frac{\partial}{\partial t}(e^{2s + 3t}) \\
\frac{\partial}{\partial s}(s \cos t) & \frac{\partial}{\partial t}(s \cos t)
\end{pmatrix}
=
\begin{pmatrix}
1 & -1 \\
2e^{2s + 3t} & 3e^{2s + 3t} \\
\cos t & -s \sin t
\end{pmatrix}
\]

At the point \( (0, 0) \), this becomes:

\[
Df(0, 0) =
\begin{pmatrix}
1 & -1 \\
2 & 3 \\
1 & 0
\end{pmatrix}
\]

### Step 4: Applying the Chain Rule

Now, using the chain rule \( D(f \circ g)(0, 0, 0) = Df(g(0, 0, 0)) \cdot Dg(0, 0, 0) \), we first compute \( g(0, 0, 0) \):

\[
g(0, 0, 0) = (0 \cdot 0 \cdot 0, 0^2 + 0^3) = (0, 0)
\]

So we need \( Df(g(0, 0, 0)) = Df(0, 0) \), which is:

\[
Df(0, 0) =
\begin{pmatrix}
1 & -1 \\
2 & 3 \\
1 & 0
\end{pmatrix}
\]

Now, multiply \( Df(0, 0) \) by \( Dg(0, 0, 0) \):

\[
D(f \circ g)(0, 0, 0) = Df(0, 0) \cdot Dg(0, 0, 0) =
\begin{pmatrix}
1 & -1 \\
2 & 3 \\
1 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
=
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\]

Thus, \( D(f \circ g)(0, 0, 0) = 0 \), which is the zero matrix.

### Final Answer:
\[
D(f \circ g)(0, 0, 0) = 0
\]

Would you like further clarification or any more details? Here are some questions to explore further:

1. What is the general formula for the chain rule in higher dimensions?
2. How do you calculate the Jacobian matrix for non-linear functions?
3. What does the zero Jacobian matrix imply about the behavior of the function at this point?
4. How does the composition of functions affect the rank of the Jacobian?
5. Can the chain rule be extended to vector-valued functions in a different manner?

**Tip:** Always verify the intermediate steps, such as computing partial derivatives and evaluating them at given points, to avoid errors when using the chain rule.

Let f(s,t) = (s - t + 2, e^{2s + 3t}, s \cos t) and g(x,y,z) = (xyz, x^2 + z^3). Use the chain rule to find D(f ∘ g)(0,0,0).

Explore how to use the chain rule to compute the derivative of a composition of functions f(s,t) = (s - t + 2, e^{2s+3t}, s cos t) and g(x,y,z) = (xyz, x^2 + z^3) at the point (0,0,0). The solution involves calculating Jacobian matrices and applying the chain rule in multivariable calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation