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Find f '(x). 4x- 1 f(x) =- 9x +5 f '(x) = Find f '(x). 40 f(x) = f '(x) = Find f '(x). 4. 20x f(x) = 9x + 14 f (x) = Find f '(x). f)=(4x-9)" (4 - x20)* f '(x) =

Find f '(x). 4x- 1 f(x) =- 9x +5 f '(x) = Find f '(x). 40 f(x) = f '(x) = Find f '(x). 4. 20x f(x) = 9x + 14 f (x) = Find f '(x). f)=(4x-9)" (4 - x20)* f '(x) =

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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**Find \( f'(x) \).** 1. \( f(x) = \frac{4x - 1}{9x + 5} + x^3 \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 2. \( f(x) = \frac{40}{x^9} \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 3. \( f(x) = \left(\frac{20x}{9x + 14}\right)^4 \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 4. \( f(x) = (4x - 9)^{14} (4 - x^{20})^4 \) \( f'(x) = \underline{\phantom{answer}} \) --- These exercises require applying different differentiation techniques such as the quotient rule, power rule, and product rule, combined where necessary.
Transcribed Image Text:**Find \( f'(x) \).** 1. \( f(x) = \frac{4x - 1}{9x + 5} + x^3 \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 2. \( f(x) = \frac{40}{x^9} \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 3. \( f(x) = \left(\frac{20x}{9x + 14}\right)^4 \) \( f'(x) = \underline{\phantom{answer}} \) --- **Find \( f'(x) \).** 4. \( f(x) = (4x - 9)^{14} (4 - x^{20})^4 \) \( f'(x) = \underline{\phantom{answer}} \) --- These exercises require applying different differentiation techniques such as the quotient rule, power rule, and product rule, combined where necessary.
Expert Solution
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.

f\left(x\right)=\frac{4x-1}{9x+5}+x^3

take derivative

f'(x)=\frac{d}{dx}\left(\frac{4x-1}{9x+5}+x^3\right)

f'(x)= \frac{d}{dx}\left(\frac{4x-1}{9x+5}\right)+\frac{d}{dx}\left(x^3\right)

f'(x)= \frac{\frac{d}{dx}\left(4x-1\right)\left(9x+5\right)-\frac{d}{dx}\left(9x+5\right)\left(4x-1\right)}{\left(9x+5\right)^2}+3x^2

f'(x)= \frac{4\left(9x+5\right)-9\left(4x-1\right)}{\left(9x+5\right)^2}+3x^2

f'(x)= \frac{36x+20-36x+9}{(9x+5)^2}+3x^2

{\color{Red} f'(x)= \frac{29}{\left(9x+5\right)^2}+3x^2}

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