If f (x) = (VT+1) (x-1 + 2) , what is f'(x)? of 'e)- (Vs +1) (-솔) + (금 + 2) (금금) o f 'a)= (VE+ 1) (P) + (금 + 3) () o f 'a)- (Vi+ 1) (구) + (1 +2) (금) +1) (교) + (금 + 2) () of to)-(VG+ 1) (¥) + (를 + 2) (₩)
If f (x) = (VT+1) (x-1 + 2) , what is f'(x)? of 'e)- (Vs +1) (-솔) + (금 + 2) (금금) o f 'a)= (VE+ 1) (P) + (금 + 3) () o f 'a)- (Vi+ 1) (구) + (1 +2) (금) +1) (교) + (금 + 2) () of to)-(VG+ 1) (¥) + (를 + 2) (₩)
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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![If \( f(x) = \left( \sqrt{x} + 1 \right) \left( x^{-1} + 2 \right) \), what is \( f'(x) \)?
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( -\frac{1}{x^2} \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( x^2 \right) + \left( \frac{1}{x} + 3 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( x^2 \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( \frac{1}{x^2} \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ef738ba-38d3-4938-bafb-9f2ae8ab4fe1%2F60896a80-23b5-4ff8-a031-bddd34c17059%2Fqh0wag_processed.png&w=3840&q=75)
Transcribed Image Text:If \( f(x) = \left( \sqrt{x} + 1 \right) \left( x^{-1} + 2 \right) \), what is \( f'(x) \)?
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( -\frac{1}{x^2} \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( x^2 \right) + \left( \frac{1}{x} + 3 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( x^2 \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
- \( f'(x) = \left( \sqrt{x} + 1 \right) \left( \frac{1}{x^2} \right) + \left( \frac{1}{x} + 2 \right) \left( \frac{1}{2\sqrt{x}} \right) \)
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