Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
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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![The expression in the image is a calculus problem that requires finding the derivative of a function defined by an integral.
**Problem Statement:**
If \( f(x) = \int_{-2}^{x^3} t^3 \, dt \) then
Find \( f'(x) \).
**Explanation:**
This is an application of the Fundamental Theorem of Calculus, specifically the Leibniz rule for differentiating integrals with variable upper limits. The theorem states that if you have a function of the form \( F(x) = \int_{a}^{g(x)} f(t) \, dt \), then the derivative \( F'(x) = f(g(x)) \cdot g'(x) \).
In this problem:
- The integral is from a constant lower bound \(-2\) to an upper bound of \(x^3\).
- The function inside the integral is \(t^3\).
Applying the theorem, differentiate:
\[ f'(x) = t^3 \Big|_{t = x^3} \cdot \frac{d}{dx}(x^3) \]
\[ f'(x) = (x^3)^3 \cdot 3x^2 \]
\[ f'(x) = x^9 \cdot 3x^2 \]
\[ f'(x) = 3x^{11} \]
Thus, the derivative of the function \( f \) with respect to \( x \) is \( 3x^{11} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6acc7099-89c3-4cde-aa7d-6eca9bf4997e%2Fa63a3d28-8fe4-4841-a5f2-f55f0feaf4f3%2Fzx1eib_processed.jpeg&w=3840&q=75)
Transcribed Image Text:The expression in the image is a calculus problem that requires finding the derivative of a function defined by an integral.
**Problem Statement:**
If \( f(x) = \int_{-2}^{x^3} t^3 \, dt \) then
Find \( f'(x) \).
**Explanation:**
This is an application of the Fundamental Theorem of Calculus, specifically the Leibniz rule for differentiating integrals with variable upper limits. The theorem states that if you have a function of the form \( F(x) = \int_{a}^{g(x)} f(t) \, dt \), then the derivative \( F'(x) = f(g(x)) \cdot g'(x) \).
In this problem:
- The integral is from a constant lower bound \(-2\) to an upper bound of \(x^3\).
- The function inside the integral is \(t^3\).
Applying the theorem, differentiate:
\[ f'(x) = t^3 \Big|_{t = x^3} \cdot \frac{d}{dx}(x^3) \]
\[ f'(x) = (x^3)^3 \cdot 3x^2 \]
\[ f'(x) = x^9 \cdot 3x^2 \]
\[ f'(x) = 3x^{11} \]
Thus, the derivative of the function \( f \) with respect to \( x \) is \( 3x^{11} \).
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