In order to find the answer to this question, we need to use the quadratic formula.

-b +- √b² - 4ac
2a

In this equation, 5 is a, 16 is b, and -84 is c. You'll notice that the formula says 'plus or minus', meaning that there are two answers depending on whether you add or subtract. Let's start by adding.

-16 + √16² -(4(5)(-84)
2(5)

Use a calculator to simplify. Remember PEMDAS (parentheses, exponents, multiplication and division, addition and subtraction) when deciding which order to go in.

-16 + 44
10
2.8

One solution is 2.8. Now, let's do the same thing, but subtract this time. Keep in mind that the number under the radical stays the same, so you don't have to recalculate that.

-16 - 44
10
-6

The solutions to the equation are 2.8 and -6.

Hope this helps!

8 0

3 years ago

Differentiate the function. y = (3x - 1)^5(4-x^4)^5

TiliK225 [7]

Answer:

$\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Distributive Property

Algebra I

Terms/Coefficients
Factoring

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 = (3x - 1)⁵(4 - x⁴)⁵

Step 2: Differentiate

Product Rule: $\displaystyle y' = \frac{d}{dx}[(3x - 1)^5](4 - x^4)^5 + (3x - 1)^5\frac{d}{dx}[(4 - x^4)^5]$
Chain Rule [Basic Power Rule]: $\displaystyle y' =[5(3x - 1)^{5-1} \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^{5-1} \cdot \frac{d}{dx}[(4 - x^4)]]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot \frac{d}{dx}[3x - 1]](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot \frac{d}{dx}[(4 - x^4)]]$
Basic Power Rule: $\displaystyle y' =[5(3x - 1)^4 \cdot 3x^{1 - 1}](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^{4-1}]$
Simplify: $\displaystyle y' =[5(3x - 1)^4 \cdot 3](4 - x^4)^5 + (3x - 1)^5[5(4 - x^4)^4 \cdot -4x^3]$
Multiply: $\displaystyle y' = 15(3x - 1)^4(4 - x^4)^5 - 20x^3(3x - 1)^5(4 - x^4)^4$
Factor: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 3(4 - x^4) - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute 3: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 4x^3(3x - 1) \bigg]$
[Distributive Property] Distribute -4x³: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4\bigg[ 12 - 3x^4 - 12x^4 + 4x^3 \bigg]$
[Brackets] Combine like terms: $\displaystyle y' = 5(3x-1)^4(4 - x^4)^4(-15x^4 + 4x^3 + 12)$
Factor: $\displaystyle y' = -5(3x-1)^4(4 - x^4)^4(15x^4 - 4x^3 - 12)$

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

Unit: Derivatives

Book: College Calculus 10e

6 0

3 years ago