How do you differentiate #f(t)= sqrt(t) / (-4t-6)# using the quotient rule?

Question

Christopher Agresta · Accepted Answer

#f'(t)=frac{2t-3}{(sqrtt)(-4t-6)^2}# #f(t)=frac{sqrtt}{-4t-6}# #f(t)=frac{t^(1/2)}{-4t-6}# Use the quotient rule to differentiate: #f'(t)=frac{(-4t-6)(1/(2t^(1/2)))-(t^(1/2))(-4)}{(-4t-6)^2}# #f'(t)=frac{(-2t-3)/(sqrtt)+4sqrtt}{(-4t-6)^2}# #f'(t)=frac{(-2t-3+4t)/(sqrtt)}{(-4t-6)^2}# #f'(t)=frac{2t-3}{(sqrtt)(-4t-6)^2}#

Adam Adkins · Answer

undefined To differentiate $ f(t) = \frac{\sqrt{t}}{-4t - 6} $ using the quotient rule:

1. Identify the numerator and denominator functions.
   - Numerator: $ \sqrt{t} $
   - Denominator: $ -4t - 6 $

2. Apply the quotient rule:
   - $ f'(t) = \frac{(denominator \cdot derivative \ of \ numerator) - (numerator \cdot derivative \ of \ denominator)}{(denominator)^2} $

3. Find the derivatives of the numerator and denominator:
   - $ \frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}} $
   - $ \frac{d}{dt}(-4t - 6) = -4 $

4. Substitute the derivatives into the quotient rule formula:
   - $ f'(t) = \frac{(-4t - 6) \cdot \frac{1}{2\sqrt{t}} - \sqrt{t} \cdot (-4)}{(-4t - 6)^2} $

5. Simplify the expression:
   - $ f'(t) = \frac{-2t - 3}{\sqrt{t} \cdot (-4t - 6)^2} $