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nadya68 [22]
3 years ago
12

Gus is six years old and has suffered intellectual and developmental disabilities, behavior problems, abnormal facial features,

and medical disorders since birth. What might Gus's mother have been exposed to during her pregnancy with Gus?
1 alcohol
2 tobacco
3 caffeine
4 excessive consumption of fish
Health
1 answer:
andre [41]3 years ago
8 0
Your answer would be 1. alcohol... Hope I helped!
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Differentiate the following functions (i) x(1+x)^3​
statuscvo [17]

Answer:

\displaystyle y' = (1 + x)^2(4x + 1)

General Formulas and Concepts:

<u>Algebra I</u>

  • Terms/Coefficients
  • Functions
  • Function Notation
  • Factoring

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative Property [Addition/Subtraction]:                                                                \displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]

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)

Explanation:

<u>Step 1: Define</u>

<em>Identify</em>

y = x(1 + x)³

<u>Step 2: Differentiate</u>

  1. Product Rule [Derivative Rule - Chain Rule]:                                                  \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot \frac{d}{dx}[1 + x]
  2. Derivative Property [Addition/Subtraction]:                                                    \displaystyle y' = \frac{d}{dx}[x] \cdot (1 + x)^3 + x \cdot \frac{d}{dx}[(1 + x)^3] \cdot (\frac{d}{dx}[1] + \frac{d}{dx}[x])
  3. Basic Power Rule:                                                                                             \displaystyle y' = x^{1 - 1} \cdot (1 + x)^3 + x \cdot 3(1 + x)^{3 - 1} \cdot (0 + x^{1 - 1})
  4. Simplify:                                                                                                             \displaystyle y' = (1 + x)^3 + 3x(1 + x)^2
  5. Factor:                                                                                                               \displaystyle y' = (1 + x)^2 \bigg[ (1 + x) + 3x \bigg]
  6. Combine like terms:                                                                                         \displaystyle y' = (1 + x)^2(4x + 1)

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

Unit: Derivatives

Book: College Calculus 10e

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