if a doctor wanted to find out the number of white blood cells in a patient’s body, he or she would order a(n) _____

Why might a patient be sent to another physician or facility?

igor_vitrenko [27]

The patient might have a Serious disease that can spread or the patient is in need of professional.

5 0

3 years ago

How do regulations differ for child care facilities compared to family daycare homes

Alenkinab [10]

A child care center cares for larger groups of children in a facility outside a private home. Child care centers may be large or small but usually divide children into groups by age, with different child care and early education professionals to work with each group. ... Neither child care setting is better than the other.

7 0

3 years ago

Replace a fuse. Correcting the source of the overload

sergejj [24]

Replace a fuse after correcting the source of the overload.

6 0

3 years ago

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