Math Problem Statement
Solution
We are given two functions and , and we are tasked with finding the derivative using the chain rule.
Step 1: Understanding the Composition of Functions
We need to compute the derivative of the composition . The chain rule for multivariable functions states that:
Where is the Jacobian matrix of , and is the Jacobian matrix of .
Step 2: Find
The Jacobian matrix is given by the partial derivatives of :
So the partial derivatives are:
\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.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Multivariable Calculus
Chain Rule
Jacobian Matrix
Formulas
Chain Rule: D(f ∘ g)(x, y, z) = Df(g(x, y, z)) ⋅ Dg(x, y, z)
Jacobian Matrix: Derivatives of vector functions
Theorems
Chain Rule for Multivariable Functions
Suitable Grade Level
Undergraduate
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