Respuesta :

Space

Answer:

[tex]\displaystyle \frac{dy}{dt} = \frac{-3}{2(4t- 3)^2\sqrt{\frac{t}{4t - 3}}}[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

Derivative Property [Multiplied Constant]:                                                           [tex]\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)[/tex]

Derivative Property [Addition/Subtraction]:                                                         [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

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

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{\frac{t}{4t - 3}}[/tex]

Step 2: Differentiate

  1. Basic Power Rule [Derivative Rule - Chain Rule]:                                       [tex]\displaystyle y' = \frac{1}{2\sqrt{\frac{t}{4t - 3}}} \cdot \frac{d}{dt} \bigg[ \frac{t}{4t - 3} \bigg][/tex]
  2. Derivative Rule [Quotient Rule]:                                                                   [tex]\displaystyle y' = \frac{1}{2\sqrt{\frac{t}{4t - 3}}} \cdot \frac{(t)'(4t - 3) - t(4t - 3)'}{(4t - 3)^2}[/tex]
  3. Basic Power Rule [Derivative Properties]:                                                   [tex]\displaystyle y' = \frac{1}{2\sqrt{\frac{t}{4t - 3}}} \cdot \frac{(4t - 3) - 4t}{(4t - 3)^2}[/tex]
  4. Simplify:                                                                                                         [tex]\displaystyle y' = \frac{-3}{2(4t- 3)^2\sqrt{\frac{t}{4t - 3}}}[/tex]

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

Unit: Differentiation

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