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1 year ago

A young person is in a public area, and an adult in a car calls out for help.

Health

1 answer:

Anarel [89]1 year ago

6 0

Hi there!

Answers are :

☆ Go over to the car to find out more about the problem -> We dont know what is their intention. So, dont do it.

☆ Offer to help the person with the problem -> Still no use. Offering doesnt help, we need to help them in action.

☆ Ask the person to come over to talk -> That's okay. Not a problem.

☆ Walk away from the person -> Never do that! It will be rude.

So, the answer is to ask the person to come over to talk.

Hope it helps !

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Derivative Notation

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Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Explanation:

Step 1: Define

Identify

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

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

Unit: Derivatives

Book: College Calculus 10e

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