Mathematics, 10.06.2020 07:57 Qwerty2771

Differentiate the function.

y = (4x − 1)^2 (4 -x^5)^4

dy/dx=

Answers: 2

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Answer from: GavinWright254

$\displaystyle y' = -4(4x - 1)(4 - x^5)^3(22x^5 - 5x^4 - 8)$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Distributive Property

Algebra I

Terms/CoefficientsFactoring

Calculus

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Step-by-step explanation:

Step 1: Define

Identify

y = (4x - 1)²(4 - x⁵)⁴

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(4x - 1)^2](4 - x^5)^4 + (4x - 1)^2\frac{d}{dx}[(4 - x^5)^4]$ Chain Rule [Basic Power Rule]: $\displaystyle y' = [2(4x - 1)^{2 - 1} \cdot \frac{d}{dx}[(4x - 1)]](4 - x^5)^4 + (4x - 1)^2[4(4 - x^5)^{4 - 1} \cdot \frac{d}{dx}[(4 - x^5)]]$ Simplify: $\displaystyle y' = [2(4x - 1) \cdot \frac{d}{dx}[(4x - 1)]](4 - x^5)^4 + (4x - 1)^2[4(4 - x^5)^3 \cdot \frac{d}{dx}[(4 - x^5)]]$ Basic Power Rule: $\displaystyle y' = [2(4x - 1) \cdot 4x^{1 - 1}](4 - x^5)^4 + (4x - 1)^2[4(4 - x^5)^3 \cdot -5x^{5 - 1}]$ Simplify: $\displaystyle y' = [2(4x - 1) \cdot 4](4 - x^5)^4 + (4x - 1)^2[4(4 - x^5)^3 \cdot -5x^4]$ Multiply: $\displaystyle y' = 8(4x - 1)(4 - x^5)^4 - 20x^4(4x - 1)^2(4 - x^5)^3$ Factor: $\displaystyle y' = 4(4x - 1)(4 - x^5)^3 \bigg[ 2(4 - x^5) - 5x^4(4x - 1) \bigg]$ [Distributive Property] Distribute 2: $\displaystyle y' = 4(4x - 1)(4 - x^5)^3 \bigg[ 8 - 2x^5 - 5x^4(4x - 1) \bigg]$ [Distributive Property] Distribute -5x⁴: $\displaystyle y' = 4(4x - 1)(4 - x^5)^3 \bigg[ 8 - 2x^5 - 20x^5 + 5x^4 \bigg]$ [Brackets] Combine like terms: $\displaystyle y' = 4(4x - 1)(4 - x^5)^3(-22x^5 + 5x^4 + 8)$ Factor: $\displaystyle y' = -4(4x - 1)(4 - x^5)^3(22x^5 - 5x^4 - 8)$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

answer: the price of one pound cabbage = $ 0.8

the point (10,8) represents that the cost of 10 pounds cabbage is $ 8

step-by-step explanation:

since, here is a proportional relation between the number of pounds of cabbage (x) and the price in dollars (y).

therefore, we can write,

$y\propto x$

⇒ y = k x

where k is the constant of proportionality.

now, point (5,4) is in the line.

thus, it will satisfies the above equation of line.

⇒ 4 = 5 k

⇒ k = 0.8

thus, the equation of the line is,

y = 0.8 x

a) for x = 1,

y = 0.8

therefore, the cost of one pound of cabbage is 0.8 dollars.

b) in the below graph (10,8) represents that the cost of 10 pounds of cabbage is 8 dollars.

Answer from: Quest

Its d so thats how it would would back from 1.0 form three four two

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Differentiate the function.

y = (4x − 1)^2 (4 -x^5)^4

dy/dx=...

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