Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 9
Verify the product law for differentiation, \((\mathbf{A B})^{\prime}=\mathbf{A}^{\prime} \mathbf{B}+\mathbf{A B}^{\prime}\). \(\mathbf{A}(t)=\left[\begin{array}{cc}t & 2 t-1 \\ t^{3} & \frac{1}{t}\end{array}\right]\) and \(\mathbf{B}(t)=\left[\begin{array}{cc}1-t & 1+t \\ 3 t^{2} & 4 t^{3}\end{array}\right]\)
Short Answer
Expert verified
The product law for differentiation is verified as both sides result in the same matrix.
Step by step solution
01
Differentiate matrix A
We need to calculate the derivative of the matrix \(\mathbf{A}(t)\). Differentiate each element of \(\mathbf{A}(t)\) with respect to \(t\). The derivative matrix \(\mathbf{A}'(t)\) is: \[\mathbf{A}'(t) = \begin{bmatrix}1 & 2 \3t^{2} & -\frac{1}{t^{2}}\end{bmatrix}\]
02
Differentiate matrix B
Next, compute the derivative of the matrix \(\mathbf{B}(t)\). Differentiate each element of \(\mathbf{B}(t)\) with respect to \(t\). The derivative matrix \(\mathbf{B}'(t)\) is: \[\mathbf{B}'(t) = \begin{bmatrix}-1 & 1 \6t & 12t^{2}\end{bmatrix}\]
03
Compute the product \(\mathbf{A}'(t)\mathbf{B}(t)\)
Multiply the differentiated matrix \(\mathbf{A}'(t)\) with the original matrix \(\mathbf{B}(t)\). Compute: \[\mathbf{A}'(t)\mathbf{B}(t) = \begin{bmatrix} 1 & 2 \ 3t^2 & -\frac{1}{t^{2}} \end{bmatrix} \begin{bmatrix} 1-t & 1+t \ 3t^2 & 4t^3 \end{bmatrix}\]After multiplication, the resulting matrix is: \[= \begin{bmatrix}1 - t + 6t^{2} & 1 + t + 8t^{3} \3t^{2}(1-t) - \frac{3t^{2}}{t^{2}} & 3t^{2}(1+t) - \frac{4t^{3}}{t^{2}}\end{bmatrix} = \begin{bmatrix}6t^{2} - t + 1 & 8t^{3} + t + 1 \3t^{2} - 3t^{3} - 3 & 3t^{2} + 3t^{3} - 4t\end{bmatrix}\]
04
Compute the product \(\mathbf{A}(t)\mathbf{B}'(t)\)
Multiply the original matrix \(\mathbf{A}(t)\) with the differentiated matrix \(\mathbf{B}'(t)\). Compute: \[\mathbf{A}(t)\mathbf{B}'(t) = \begin{bmatrix} t & 2t-1 \ t^{3} & \frac{1}{t} \end{bmatrix} \begin{bmatrix} -1 & 1 \ 6t & 12t^{2} \end{bmatrix}\]After multiplication, the resulting matrix is: \[= \begin{bmatrix}-1t + 12t^{2} - 6t + 1 & 12t^{3} \-1t^{3} + 6t^{2} & 12t^{6} + 12\end{bmatrix} = \begin{bmatrix}12t^{2} - 7t + 1 & 12t^{3} \-1t^{3} + 6t^{2} & 12t^{6} + 12\end{bmatrix}\]
05
Add results from Steps 3 and 4
Add the two results obtained from the products \(\mathbf{A}'(t)\mathbf{B}(t)\) and \(\mathbf{A}(t)\mathbf{B}'(t)\): \[\begin{bmatrix}6t^2 - t + 1 & 8t^3 + t + 1 \3t^2 - 3t^3 - 3 & 3t^2 + 3t^3 - 4t\end{bmatrix} + \begin{bmatrix}12t^2 - 7t + 1 & 12t^3 \-t^3 + 6t^2 & 12t^6 + 12\end{bmatrix}\]The resulting matrix is: \[\begin{bmatrix}18t^2 - 8t + 2 & 20t^3 + t + 1 \2t^2 - 4t^3 - 3 & 12t^6 + 3t^3 - 4t + 12\end{bmatrix}\]
06
Differentiate the product \((\mathbf{A}(t)\mathbf{B}(t))\) and verify
Calculate the product \(\mathbf{A}(t)\mathbf{B}(t)\):\[\mathbf{A}(t)\mathbf{B}(t) = \begin{bmatrix}t(1-t) + (2t-1)(3t^2) & t(1+t) + (2t-1)(4t^3) \t^3(1-t) + \frac{1}{t}(3t^2) & t^3(1+t) + \frac{1}{t}(4t^3)\end{bmatrix}\]Simplify to: \[\begin{bmatrix}t - t^2 + 6t^3 - 3t^2 & t + t^2 + 8t^3 - 4t^3\t^3 - t^4 + 3t & t^3 + t^4 + 4t^2\end{bmatrix} = \begin{bmatrix} -2t^2 + 6t^3 + t & 6t^3 + t + t^2 \ -t^4 + t^3 + 3t & t^4 + t^3 + 4t^2 \ \end{bmatrix}\]Differentiate this result and verify it equals the matrix from Step 5 to confirm the product rule. The derivative:\[\begin{bmatrix}18t^2 - 8t + 2 & 20t^3 + t + 1 \2t^2 - 4t^3 - 3 & 12t^6 + 3t^3 - 4t + 12\end{bmatrix}.\]
07
Verification
Both calculations yield the same resulting matrix, confirming that the product rule holds for these matrices.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Product Rule for Differentiation
In calculus, the product rule is crucial when differentiating expressions that are products of two functions. For matrices, this rule transforms into the equation \((\mathbf{A B})^{\prime}=\mathbf{A}^{\prime} \mathbf{B}+\mathbf{A B}^{\prime}\). This means that to find the derivative of the product of two matrices \(\mathbf{A}(t)\) and \(\mathbf{B}(t)\), you need to:
- First differentiate \(\mathbf{A}(t)\), and then multiply the resultant matrix by \(\mathbf{B}(t)\).
- Next, differentiate \(\mathbf{B}(t)\) and multiply this derivative matrix by the original \(\mathbf{A}(t)\).
- Finally, add these two results together.
Matrix Differentiation
Matrix differentiation is the process of computing the derivative of a matrix whose elements are functions of a variable, often \(t\). Just like differentiating a function, here, each element of the matrix is differentiated individually.
- To differentiate a matrix like \(\mathbf{A}(t)\), you perform the differentiation element-wise. Consider each element as a separate small function.
- For example, if an element is \(t^3\), its derivative is \(3t^2\).
Matrix Multiplication
Matrix multiplication is a bit different from regular multiplication. Here, for two matrices to be multiplied, certain criteria need to be met. The number of columns in the first matrix must match the number of rows in the second matrix.
- To multiply two matrices \(\mathbf{A}\) and \(\mathbf{B}\), each element of the resulting matrix is calculated as the dot product of the corresponding row of \(\mathbf{A}\) and column of \(\mathbf{B}\).
- This involves multiplying corresponding elements and summing up those products.
