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Find the first and second derivative of the function. G(r) = √r + √√√r G'(r) G"(r) || = 1 1 + (²) 2√r 6r 1 4r NIG 8 17 (¹7) 36r

Find the first and second derivative of the function. G(r) = √r + √√√r G'(r) G"(r) || = 1 1 + (²) 2√r 6r 1 4r NIG 8 17 (¹7) 36r

Calculus: Early Transcendentals
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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### Finding the First and Second Derivative of the Function Consider the function \( G(r) = \sqrt{r} + \sqrt[6]{r} \). #### First Derivative To find the first derivative \( G'(r) \), we differentiate each term of the function with respect to \( r \): \[ G'(r) = \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] Thus, the first derivative of \( G(r) \) is: \[ G'(r) = \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] #### Second Derivative To find the second derivative \( G''(r) \), we further differentiate \( G'(r) \) with respect to \( r \): \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] Simplifying the constants: \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] Therefore, the second derivative of \( G(r) \) is: \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] #### Summary 1. The first derivative \( G'(r) \) of the function \( G(r) = \sqrt{r} + \sqrt[6]{r} \) is \[ \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] 2. The second derivative \( G''(r) \) of the same function is \[ -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}}. \]
Transcribed Image Text:### Finding the First and Second Derivative of the Function Consider the function \( G(r) = \sqrt{r} + \sqrt[6]{r} \). #### First Derivative To find the first derivative \( G'(r) \), we differentiate each term of the function with respect to \( r \): \[ G'(r) = \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] Thus, the first derivative of \( G(r) \) is: \[ G'(r) = \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] #### Second Derivative To find the second derivative \( G''(r) \), we further differentiate \( G'(r) \) with respect to \( r \): \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] Simplifying the constants: \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] Therefore, the second derivative of \( G(r) \) is: \[ G''(r) = -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}} \] #### Summary 1. The first derivative \( G'(r) \) of the function \( G(r) = \sqrt{r} + \sqrt[6]{r} \) is \[ \frac{1}{6r^{\left(\frac{8}{9}\right)}} + \frac{1}{2\sqrt{r}} \] 2. The second derivative \( G''(r) \) of the same function is \[ -\frac{1}{4r^{\left(\frac{3}{2}\right)}} - \frac{8}{36r^{\left(\frac{17}{9}\right)}}. \]
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