Question

Derivative of the (sqrt of t/4t-3)

Answer 1

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

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

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

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

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

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Explanation:

Step 1: Define

Identify

`\displaystyle y = \sqrt{(t)/(4t - 3)}`

Step 2: Differentiate

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

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

Unit: Differentiation