Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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**Calculus Problem: Differential Calculus** **Problem Statement:** Given \( F(x) \) below, find \( F'(x) \). \[ F(x) = \int_{2}^{\cos(x)} (t^2 - 2) \, dt \] **Instructions:** Provide your answer below: \[ F'(x) = \boxed{} \] --- ### Explanation: To solve this problem, we need to find the derivative of \( F(x) \) with respect to \( x \). The function \( F(x) \) is defined as an integral with variable upper limits, and we can apply the Leibniz rule for differentiation under the integral sign in this scenario. Leibniz's rule states that if we have an integral of the form: \[ G(x) = \int_{a(x)}^{b(x)} f(t) \, dt \] Then, the derivative \( G'(x) \) is given by: \[ G'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \] In our specific problem, the lower limit \( a(x) = 2 \) is a constant, so \( a'(x) = 0 \), and the upper limit \( b(x) = \cos(x) \), so \( b'(x) = -\sin(x) \). Given: \[ f(t) = t^2 - 2 \] We apply the rule: \[ F'(x) = (t^2 - 2) \bigg|_{t=\cos(x)} \cdot \frac{d}{dx}[\cos(x)] \] Evaluating \( f(t) \) at the upper limit \( t = \cos(x) \): \[ f(\cos(x)) = (\cos(x))^2 - 2 \] Then: \[ F'(x) = ((\cos(x))^2 - 2) \cdot (-\sin(x)) \] So, the final answer is: \[ F'(x) = -((\cos(x))^2 - 2) \sin(x) \]
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Transcribed Image Text:**Calculus Problem: Differential Calculus** **Problem Statement:** Given \( F(x) \) below, find \( F'(x) \). \[ F(x) = \int_{2}^{\cos(x)} (t^2 - 2) \, dt \] **Instructions:** Provide your answer below: \[ F'(x) = \boxed{} \] --- ### Explanation: To solve this problem, we need to find the derivative of \( F(x) \) with respect to \( x \). The function \( F(x) \) is defined as an integral with variable upper limits, and we can apply the Leibniz rule for differentiation under the integral sign in this scenario. Leibniz's rule states that if we have an integral of the form: \[ G(x) = \int_{a(x)}^{b(x)} f(t) \, dt \] Then, the derivative \( G'(x) \) is given by: \[ G'(x) = f(b(x)) \cdot b'(x) - f(a(x)) \cdot a'(x) \] In our specific problem, the lower limit \( a(x) = 2 \) is a constant, so \( a'(x) = 0 \), and the upper limit \( b(x) = \cos(x) \), so \( b'(x) = -\sin(x) \). Given: \[ f(t) = t^2 - 2 \] We apply the rule: \[ F'(x) = (t^2 - 2) \bigg|_{t=\cos(x)} \cdot \frac{d}{dx}[\cos(x)] \] Evaluating \( f(t) \) at the upper limit \( t = \cos(x) \): \[ f(\cos(x)) = (\cos(x))^2 - 2 \] Then: \[ F'(x) = ((\cos(x))^2 - 2) \cdot (-\sin(x)) \] So, the final answer is: \[ F'(x) = -((\cos(x))^2 - 2) \sin(x) \]
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