Let \( f(x) = \frac{1}{2x + 1} \). Use the derivative definition: \( f'(x) = \lim_{h \to 0} \frac{f(x + h) – f(x)}{h} \).

\[ f(x + h) = \frac{1}{2(x + h) + 1} = \frac{1}{2x + 2h + 1} \]

\[ f'(x) = \lim_{h \to 0} \frac{\frac{1}{2x + 2h + 1} – \frac{1}{2x + 1}}{h} \]

Common denominator: \( (2x + 2h + 1)(2x + 1) \).

\[ \frac{\frac{1}{2x + 2h + 1} – \frac{1}{2x + 1}}{h} = \frac{(2x + 1) – (2x + 2h + 1)}{h (2x + 2h + 1)(2x + 1)} = \frac{-2h}{h (2x + 2h + 1)(2x + 1)} = \frac{-2}{(2x + 2h + 1)(2x + 1)} \]

Take the limit: \[ \lim_{h \to 0} \frac{-2}{(2x + 2h + 1)(2x + 1)} = \frac{-2}{(2x + 1)(2x + 1)} = \frac{-2}{(2x + 1)^2} \]

Answer: \( f'(x) = \frac{-2}{(2x + 1)^2} \), as shown.

Prove by induction: \( \frac{d^n}{dx^n} \left( (2x + 1)^{-1} \right) = (-1)^n \frac{2^n n!}{(2x + 1)^{n + 1}} \).

Base case (\( n = 1 \)):

\[ \frac{d}{dx} \left( \frac{1}{2x + 1} \right) = (-1)^1 \frac{2^1 \cdot 1!}{(2x + 1)^{1 + 1}} = \frac{-2}{(2x + 1)^2} \]

From part (a), this is true.

Induction hypothesis: Assume true for \( n = k \): \[ \frac{d^k}{dx^k} \left( (2x + 1)^{-1} \right) = (-1)^k \frac{2^k k!}{(2x + 1)^{k + 1}} \]

Induction step (\( n = k + 1 \)):

Differentiate: \[ \frac{d^{k + 1}}{dx^{k + 1}} = \frac{d}{dx} \left( (-1)^k \frac{2^k k!}{(2x + 1)^{k + 1}} \right) = (-1)^k 2^k k! \frac{d}{dx} \left( (2x + 1)^{-k – 1} \right) \]

\[ \frac{d}{dx} \left( (2x + 1)^{-k – 1} \right) = (-k – 1) (2x + 1)^{-k – 2} \cdot 2 = -2 (k + 1) (2x + 1)^{-k – 2} \]

\[ \frac{d^{k + 1}}{dx^{k + 1}} = (-1)^k 2^k k! \cdot \left( -2 (k + 1) (2x + 1)^{-k – 2} \right) = (-1)^{k + 1} 2^{k + 1} (k + 1) k! (2x + 1)^{-k – 2} \]

\[ = (-1)^{k + 1} \frac{2^{k + 1} (k + 1)!}{(2x + 1)^{k + 2}} \]

True for \( n = k + 1 \). Since true for \( n = 1 \), the result holds by induction.

Answer: \( \frac{d^n}{dx^n} \left( (2x + 1)^{-1} \right) = (-1)^n \frac{2^n n!}{(2x + 1)^{n + 1}} \), as shown.