In Exercises 65 and 66, find the derivative using the definition.


ƒ(t) = 1 .
2t + 1

Verified step by step guidance

1. Start by recalling the definition of the derivative: the derivative of a function \( f(t) \) at a point \( t \) is given by \( f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \).
2. Identify the function \( f(t) = \frac{1}{2t + 1} \). We need to find \( f(t+h) \), which is \( \frac{1}{2(t+h) + 1} = \frac{1}{2t + 2h + 1} \).
3. Substitute \( f(t) \) and \( f(t+h) \) into the derivative definition: \( f'(t) = \lim_{h \to 0} \frac{\frac{1}{2t + 2h + 1} - \frac{1}{2t + 1}}{h} \).
4. To simplify the expression, find a common denominator for the fractions in the numerator: \( (2t + 2h + 1)(2t + 1) \). Rewrite the expression as \( \frac{(2t + 1) - (2t + 2h + 1)}{h(2t + 2h + 1)(2t + 1)} \).
5. Simplify the numerator: \( (2t + 1) - (2t + 2h + 1) = -2h \). The expression becomes \( \lim_{h \to 0} \frac{-2h}{h(2t + 2h + 1)(2t + 1)} \). Cancel \( h \) from the numerator and denominator, then evaluate the limit as \( h \to 0 \).