Given F(x) below, find F'(x). cos - Las () (1² - 2) dr F(x) =

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...
icon
Related questions
Question
He1p7!!???
**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:

\[ F'(x) = -((\cos(x))^2 - 2) \sin(x) \]
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: \[ F'(x) = -((\cos(x))^2 - 2) \sin(x) \]
Expert Solution
steps

Step by step

Solved in 2 steps with 1 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning