1. Use the Fundamental Theorem of Calculus,, to find each derivative: (a) 13 d²f² √9 - y² dy dx x

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
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## Problem 1: Using the Fundamental Theorem of Calculus to Find Derivatives

### (a)

\[ 
\frac{d}{dx} \int_{x}^{13} \sqrt{9 - y^2} \, dy 
\]

### (b)

\[ 
\frac{d}{dx} \int_{0}^{3^{\tan x}} \frac{9 + y^2}{2y + 1} \, dy 
\]

### (c)

\[ 
\frac{d}{dx} \int_{2\sqrt{x}}^{3^{x^3}} (1 + t) \, dy 
\]

**Explanation:**

- **Part (a):** The integral's limits of integration are from \(x\) to 13. The function inside the integral is \(\sqrt{9 - y^2}\).
  
- **Part (b):** The limits are from 0 to \(3^{\tan x}\), with the integrand \(\frac{9 + y^2}{2y + 1}\).

- **Part (c):** The limits are from \(2\sqrt{x}\) to \(3^{x^3}\) and the integrand is \(1 + t\).

Each problem involves differentiating a definite integral with respect to \(x\), applying the Fundamental Theorem of Calculus, which relates differentiation and integration.
Transcribed Image Text:## Problem 1: Using the Fundamental Theorem of Calculus to Find Derivatives ### (a) \[ \frac{d}{dx} \int_{x}^{13} \sqrt{9 - y^2} \, dy \] ### (b) \[ \frac{d}{dx} \int_{0}^{3^{\tan x}} \frac{9 + y^2}{2y + 1} \, dy \] ### (c) \[ \frac{d}{dx} \int_{2\sqrt{x}}^{3^{x^3}} (1 + t) \, dy \] **Explanation:** - **Part (a):** The integral's limits of integration are from \(x\) to 13. The function inside the integral is \(\sqrt{9 - y^2}\). - **Part (b):** The limits are from 0 to \(3^{\tan x}\), with the integrand \(\frac{9 + y^2}{2y + 1}\). - **Part (c):** The limits are from \(2\sqrt{x}\) to \(3^{x^3}\) and the integrand is \(1 + t\). Each problem involves differentiating a definite integral with respect to \(x\), applying the Fundamental Theorem of Calculus, which relates differentiation and integration.
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