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Elenna [48]
2 years ago
8

Pls Help - Calc. HW dy/dx problem

Mathematics
1 answer:
GREYUIT [131]2 years ago
8 0

Answer:

\displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}

General Formulas and Concepts:

<u>Calculus</u>

Differentiation

  • 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)]

Derivative Rule [Basic Power Rule]:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:                                                                           \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify.</em>

\displaystyle y = \frac{\log (x)}{\log (x) - 2}

<u>Step 2: Differentiate</u>

  1. [Function] Derivative Rule [Quotient Rule]:                                                 \displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x) - 2]'[\log (x)]}{[\log (x) - 2]^2}
  2. Rewrite [Derivative Rule - Addition/Subtraction]:                                       \displaystyle y' = \frac{[\log (x) - 2][\log (x)]' - [\log (x)' - 2'][\log (x)]}{[\log (x) - 2]^2}
  3. Logarithmic Differentiation:                                                                         \displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - [\frac{1}{\ln (10)x} - 2'][\log (x)]}{[\log (x) - 2]^2}
  4. Derivative Rule [Basic Power Rule]:                                                             \displaystyle y' = \frac{[\log (x) - 2]\frac{1}{\ln (10)x} - \frac{1}{\ln (10)x}[\log (x)]}{[\log (x) - 2]^2}
  5. Simplify:                                                                                                         \displaystyle y' = \frac{\frac{\log (x) - 2}{\ln (10)x} - \frac{\log (x)}{\ln (10)x}}{[\log (x) - 2]^2}
  6. Simplify:                                                                                                         \displaystyle y' = \frac{\frac{-2}{\ln (10)x}}{[\log (x) - 2]^2}
  7. Rewrite:                                                                                                         \displaystyle y' = \frac{-2}{x \ln (10)[\log (x) - 2]^2}

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

Unit: Differentiation

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Question 23:

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Question 24:

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Step-by-step explanation:

As RS is the perpendicular bisector of DE, it will divide DE in two equal parts DS and SE

<u>Question number 23:</u>

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Question 23:

x = 4

DE = 44

Question 24:

x = 25

SE = 28

Keywords: Bisector, Line segment

Learn more about line segments at:

  • brainly.com/question/629998
  • brainly.com/question/6208262

#LearnwithBrainly

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