Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Concept explainers

Rate of Change
The relation between two quantities which displays how much greater one quantity is than another is called ratio.
Slope
The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:
Question

Help on where to start or answer possibly. Do I start with finding g'(x)?

The image contains a mathematical problem involving definite integrals and function differentiation. The problem is stated as follows: "If \( f(x) = \int_{0}^{g(x)} \frac{1}{\sqrt{1 + t^3}} \, dt \), where \( g(x) = \int_{0}^{\cos(x)} [1 + \sin(t^2)] \, dt \), find \( f'\left(\frac{\pi}{2}\right) \)." This problem asks for the derivative of the function \( f(x) \) evaluated at \( x = \frac{\pi}{2} \). The function \( f(x) \) is given as an integral with a variable upper limit defined by another integral \( g(x) \). The challenge involves applying techniques such as differentiation under the integral sign and the Fundamental Theorem of Calculus.
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Transcribed Image Text:The image contains a mathematical problem involving definite integrals and function differentiation. The problem is stated as follows: "If \( f(x) = \int_{0}^{g(x)} \frac{1}{\sqrt{1 + t^3}} \, dt \), where \( g(x) = \int_{0}^{\cos(x)} [1 + \sin(t^2)] \, dt \), find \( f'\left(\frac{\pi}{2}\right) \)." This problem asks for the derivative of the function \( f(x) \) evaluated at \( x = \frac{\pi}{2} \). The function \( f(x) \) is given as an integral with a variable upper limit defined by another integral \( g(x) \). The challenge involves applying techniques such as differentiation under the integral sign and the Fundamental Theorem of Calculus.
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