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Olin [163]
2 years ago
11

June is driving from Brookline, Massachusetts to Brooklyn, new york. The cities are 200 miles apart. Pretend June lives in a wor

ld in which she encounters no traffic jams and can drive at a constant speed of 50mph the entire way. Consider the following function. the distance June has traveled, c, is a function of her driving time, t.
Mathematics
1 answer:
I am Lyosha [343]2 years ago
4 0

It took June 4 hours to get from Brookline to Brooklyn.

<h3><u>Time calculation</u></h3>

Given that June is driving from Brookline, Massachusetts to Brooklyn, New York, and the cities are 200 miles apart, and she can drive at a constant speed of 50mph the entire way, to determine the function that solves the time June takes to complete the route, the following calculation must be made:

  • C / Constant speed = T
  • 200 / 50 = T
  • 4 = T

So it took June 4 hours to get from Brookline to Brooklyn.

Learn more about time calculation in brainly.com/question/26707195

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Step-by-step explanation:

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Find the derivative of following function.
Aleks04 [339]

Answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

General Formulas and Concepts:
<u>Calculus</u>

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:
\displaystyle (cu)' = cu'

Derivative Property [Addition/Subtraction]:
\displaystyle (u + v)' = u' + v'

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]:
\displaystyle (uv)' = u'v + uv'

Derivative Rule [Quotient Rule]:
\displaystyle \bigg( \frac{u}{v} \bigg)' = \frac{vu' - uv'}{v^2}

Derivative Rule [Chain Rule]:
\displaystyle [u(v)]' = u'(v)v'

Step-by-step explanation:

*Note:

Since the answering box is <em>way</em> too small for this problem, there will be limited explanation.

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle y = \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \times \frac{\tan^2 x + 5x}{\csc^2 x + 3}

<u>Step 2: Differentiate</u>

We can differentiate this function with the use of the given <em>derivative rules and properties</em>.

Applying Product Rule:

\displaystyle y' = \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) '

Differentiating the first portion using Quotient Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( \cos^2 x - 3\sqrt{x} + 6 \big)' \big( \sin^2 x + 6 \big) - \big( \sin^2 x + 6 \big)' \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule:

\displaystyle \bigg( \frac{\cos^2 x - 3\sqrt{x} + 6}{\sin^2 x + 6} \bigg)' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2}

Differentiating the second portion using Quotient Rule again:

\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( \tan^2 x + 5x \big)' \big( \csc^2 x + 3 \big) - \big( \csc^2 x + 3 \big)' \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Apply Derivative Rules and Properties, namely the Chain Rule again:
\displaystyle \bigg( \frac{\tan^2 x + 5x}{\csc^2 x + 3} \bigg) ' = \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Substitute in derivatives:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - \big( 2 \sin x \cos x \big) \big( \cos^2 x - 3\sqrt{x} + 6 \big)}{\big( \sin^2 x + 6 \big)^2} \frac{\tan^2 x + 5x}{\csc^2 x + 3} + \frac{\cos^2 x - 3\sqrt{x} +6}{\sin^2 x + 6} \frac{\big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) - \big( -2 \csc^2 x \cot x \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2}

Simplify:

\displaystyle y' = \frac{\big( \tan^2 x + 5x \big) \bigg[ \big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \sin^2 x + 6 \big) - 2 \sin x \cos x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \bigg]}{\big( \sin^2 x + 6 \big)^2 \big( \csc^2 x + 3 \big)} + \frac{\big( \cos^2 x - 3\sqrt{x} +6 \big) \bigg[ \big( 2 \tan x \sec^2 x + 5 \big) \big( \csc^2 x + 3 \big) + 2 \csc^2 x \cot x \big( \tan^2 x + 5x \big) \bigg] }{\big( \csc^2 x + 3 \big)^2 \big( \sin^2 x + 6 \big)}

We can rewrite the differential by factoring and common mathematical properties to obtain our final answer:

\displaystyle y' = \frac{\big( -2 \cos x \sin x - \frac{3}{2\sqrt{x}} \big) \big( \tan^2 x + 5x \big) + \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( 2 \sec^2 x \tan x + 5 \big)}{ \big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)} + \frac{2 \cot x \csc^2 x \big( \cos^2 x - 3\sqrt{x} + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big)^2 \big( \sin^2x + 6 \big)} - \frac{2 \cos x \sin x \big( \cos^2 x - 3\sqrt{x}  + 6 \big) \big( \tan^2 x + 5x \big)}{\big( \csc^2 x + 3 \big) \big( \sin^2 x + 6 \big)^2}

∴ we have found our derivative.

---

Learn more about derivatives: brainly.com/question/26836290

Learn more about calculus: brainly.com/question/23558817

---

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

Unit: Differentiation

8 0
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