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3 years ago

5

When reading the level of a fluid in a graduated cylinder, what part of the meniscus is lined up with the graduations to give an

accurate reading of the volume?"

1 answer:

Lapatulllka [165]3 years ago

8 0

The bottom part and not the edges of the <span>meniscus</span>.

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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>

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