Verify the product law for differentiation, (AB)' = A'B + AB' where A(t) = To calculate (AB)', first calculate AB. 10t³-8t² + 3t 10t - 5t² + 3t² + 3t AB= 4 3 -t+t+15t t² +t² +15t² Now take the derivative of AB to find (AB)'. (AB)' = 3t 2t-1 3 and B(t) = 5t² 5t³
Verify the product law for differentiation, (AB)' = A'B + AB' where A(t) = To calculate (AB)', first calculate AB. 10t³-8t² + 3t 10t - 5t² + 3t² + 3t AB= 4 3 -t+t+15t t² +t² +15t² Now take the derivative of AB to find (AB)'. (AB)' = 3t 2t-1 3 and B(t) = 5t² 5t³
Calculus: Early Transcendentals
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ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Verification of the Product Law for Differentiation
#### Problem Statement
Verify the product law for differentiation, given by the equation:
\[ (AB)' = A'B + AB' \]
where \( A(t) \) and \( B(t) \) are defined as:
\[ A(t) = \begin{bmatrix} 3t & 2t - 1 \\ t^3 & \frac{3}{t} \end{bmatrix} \]
and
\[ B(t) = \begin{bmatrix} 1 - t & 1 + t \\ 5t^2 & 5t^3 \end{bmatrix} \]
#### Steps to Solution
1. **Calculate the Product \( AB \)**
To calculate \((AB)'\), first, we need to find \( AB \).
Using the given matrices:
\[ AB = \begin{bmatrix} 3t & 2t - 1 \\ t^3 & \frac{3}{t} \end{bmatrix} \begin{bmatrix} 1 - t & 1 + t \\ 5t^2 & 5t^3 \end{bmatrix} \]
Perform the matrix multiplication:
\[ AB = \begin{bmatrix} (3t)(1-t) + (2t-1)(5t^2) & (3t)(1+t) + (2t-1)(5t^3) \\ (t^3)(1-t) + \left(\frac{3}{t}\right)(5t^2) & (t^3)(1+t) + \left(\frac{3}{t}\right)(5t^3) \end{bmatrix} \]
Breaking down each element:
\[ AB = \begin{bmatrix} 3t - 3t^2 + 10t^3 - 5t^2 & 3t + 3t^2 + 10t^4 - 5t^3 \\ t^3 - t^4 + 15t & t^3 + t^4 + 15t^2 \end{bmatrix} \]
Simplifying further, we get:
\[ AB = \begin{bmatrix} 10t^3 - 8t^2 + 3t & 10t^4 - 5t](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F06bbd155-c9d1-4bdc-8881-0cd893ac016a%2Ff4b1cf86-4ce1-4d20-b583-48490adaf4c1%2Faxqy7f9_processed.png&w=3840&q=75)
Transcribed Image Text:### Verification of the Product Law for Differentiation
#### Problem Statement
Verify the product law for differentiation, given by the equation:
\[ (AB)' = A'B + AB' \]
where \( A(t) \) and \( B(t) \) are defined as:
\[ A(t) = \begin{bmatrix} 3t & 2t - 1 \\ t^3 & \frac{3}{t} \end{bmatrix} \]
and
\[ B(t) = \begin{bmatrix} 1 - t & 1 + t \\ 5t^2 & 5t^3 \end{bmatrix} \]
#### Steps to Solution
1. **Calculate the Product \( AB \)**
To calculate \((AB)'\), first, we need to find \( AB \).
Using the given matrices:
\[ AB = \begin{bmatrix} 3t & 2t - 1 \\ t^3 & \frac{3}{t} \end{bmatrix} \begin{bmatrix} 1 - t & 1 + t \\ 5t^2 & 5t^3 \end{bmatrix} \]
Perform the matrix multiplication:
\[ AB = \begin{bmatrix} (3t)(1-t) + (2t-1)(5t^2) & (3t)(1+t) + (2t-1)(5t^3) \\ (t^3)(1-t) + \left(\frac{3}{t}\right)(5t^2) & (t^3)(1+t) + \left(\frac{3}{t}\right)(5t^3) \end{bmatrix} \]
Breaking down each element:
\[ AB = \begin{bmatrix} 3t - 3t^2 + 10t^3 - 5t^2 & 3t + 3t^2 + 10t^4 - 5t^3 \\ t^3 - t^4 + 15t & t^3 + t^4 + 15t^2 \end{bmatrix} \]
Simplifying further, we get:
\[ AB = \begin{bmatrix} 10t^3 - 8t^2 + 3t & 10t^4 - 5t
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