Answer: Let \(P_n\) be the statement that \(\frac{d^n}{dx^n}\left(\frac{1}{2x+3}\right)=(-1)^n\frac{n!2^n}{(2x+3)^{n+1}}\).

For \(n=1\),

\(\frac{d}{dx}\left(\frac{1}{2x+3}\right)=\frac{d}{dx}(2x+3)^{-1}=-1(2x+3)^{-2}\cdot 2=-\frac{2}{(2x+3)^2}\).

This is equal to \((-1)^1\frac{1!\,2^1}{(2x+3)^{1+1}}\), so \(P_1\) is true.

Assume that \(P_k\) is true for some positive integer \(k\), so that

\(\frac{d^k}{dx^k}\left(\frac{1}{2x+3}\right)=(-1)^k\frac{k!2^k}{(2x+3)^{k+1}}\).

Differentiate both sides with respect to \(x\):

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^k k!2^k\frac{d}{dx}\left((2x+3)^{-(k+1)}\right)\).

Using the chain rule,

\(\frac{d}{dx}\left((2x+3)^{-(k+1)}\right)=-(k+1)(2x+3)^{-(k+2)}\cdot 2\).

Hence

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^k k!2^k\bigl[-2(k+1)(2x+3)^{-(k+2)}\bigr]\).

So

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^{k+1}(k+1)!2^{k+1}(2x+3)^{-(k+2)}\)

\(= (-1)^{k+1}\frac{(k+1)!2^{k+1}}{(2x+3)^{k+2}}\).

Therefore \(P_{k+1}\) is true whenever \(P_k\) is true.

Since \(P_1\) is true and \(P_k\Rightarrow P_{k+1}\), the result holds for all positive integers \(n\).

Let \(P_n\) be the proposition

\(\frac{d^n}{dx^n}\left(\frac{1}{2x+3}\right)=(-1)^n\frac{n!2^n}{(2x+3)^{n+1}}\).

We prove \(P_n\) by induction.

For \(n=1\),

\(\frac{d}{dx}\left(\frac{1}{2x+3}\right)=\frac{d}{dx}(2x+3)^{-1}=-1(2x+3)^{-2}\cdot 2=-\frac{2}{(2x+3)^2}\).

Also, when \(n=1\), the formula gives

\((-1)^1\frac{1!2^1}{(2x+3)^{1+1}}=-\frac{2}{(2x+3)^2}\).

So \(P_1\) is true.

Now assume \(P_k\) is true for some positive integer \(k\), so

\(\frac{d^k}{dx^k}\left(\frac{1}{2x+3}\right)=(-1)^k\frac{k!2^k}{(2x+3)^{k+1}}\).

Differentiate both sides with respect to \(x\):

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^k k!2^k\frac{d}{dx}\left((2x+3)^{-(k+1)}\right)\).

By the chain rule,

\(\frac{d}{dx}\left((2x+3)^{-(k+1)}\right)=-(k+1)(2x+3)^{-(k+2)}\cdot 2\).

Hence

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^k k!2^k\cdot \bigl[-2(k+1)(2x+3)^{-(k+2)}\bigr]\).

Therefore

\(\frac{d^{k+1}}{dx^{k+1}}\left(\frac{1}{2x+3}\right)=(-1)^{k+1}(k+1)!2^{k+1}(2x+3)^{-(k+2)}\)

\(= (-1)^{k+1}\frac{(k+1)!2^{k+1}}{(2x+3)^{k+2}}\).

So \(P_{k+1}\) is true whenever \(P_k\) is true. Therefore, by mathematical induction,

\(\frac{d^n}{dx^n}\left(\frac{1}{2x+3}\right)=(-1)^n\frac{n!2^n}{(2x+3)^{n+1}}\)

for all positive integers \(n\).