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IgorC [24]
3 years ago
8

What language do the medical terms come from?

Health
2 answers:
madam [21]3 years ago
8 0
They come from Ancient Greek
nignag [31]3 years ago
6 0
They come from latin
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Differentiate the following functions (i) x(1+x)^3​
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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

5 0
3 years ago
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