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3 years ago

Is the coxal region inferior to the abdomen

Health

1 answer:

leonid [27]3 years ago

6 0

Answer:

It would be inferior to the abdomen because the coxal region is the belt line.

Explanation:

The coxal region encompassing the hip area. the pubic region encompassing the area above the genitals.Nov

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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:

Algebra I

Terms/Coefficients
Functions
Function Notation
Factoring

Calculus

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:

Step 1: Define

Identify

y = x(1 + x)³

Step 2: Differentiate

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