I'm trying to solve this following question:
By deriving $f(x) = x^n (1+x)^n$ $n$ times, determine the value of $\sum_{k=0}^n\binom{n}{k}^2$.
My attempt on this was to express $f(x)$ in 2 ways: $f(x)=x^n(1+x)^n$ and $$ f(x) = x^n \sum_{k=0}^n \binom{n}{k}x^k = \sum_{k=0}^n \binom{n}{k} x^{n+k}. $$
By deriving both expressions we find: $$ f^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} \frac{n!}{(n-k)!} x^{n-k} \times \frac{n!}{k!} (1+x)^{k} = n! \sum_{k=0}^n \binom{n}{k}^2 x^{n-k}(1+x)^k $$ and $$ f^{(n)}(x) = \sum_{k=0}^n \binom{n}{k} \frac{(n+k)!}{k!} x^k. $$ Hence for $x=1$: $$ n! \sum_{k=0}^n \binom{n}{k}^2 = \sum_{k=0}^n \binom{n}{k} \frac{(n+k)!}{k!} $$
Hence, $$ \sum_{k=0}^n \binom{n}{k}^2 = \sum_{k=0}^n \binom{n}{k} \binom{n+k}{n}. $$ This is where I'm stuck as I don't know how to evaluate this sum.
The expression: $$\sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$$ Is well known to be a special case of vandermonde's identity.
To do this with a derivative of $f(x) = (1+x)^n x^n$ we use leibeniz's identity (note that $f^{(k)}$ is shorthand for the $k^th$ derivative of $f$ and $x^{\underline{k}}$ is the falling factorial function): $$\begin{align}f^{(n)}(x) &= \sum_{k=0}^n {n \choose k} (x^n)^{(k)} \left((1+x)^n \right)^{(n-k)} \\&= \sum_{k=0}^n {n \choose k} n^{\underline{k}} x^{n-k} n^{\underline{n-k}} (1+x)^{k}\\&=\sum_{k=0}^n {n \choose k}^2 n! x^{n-k} (1+x)^{k} \end{align}$$
But we also know that $x^n(1+x)^n = \sum_{k=0}^n {n \choose k} x^{n+k}$ And of course the $n^{th}$ derivative comes out to
$$\sum_{k=0}^n (n+k)^{\underline{k}} {n \choose k} x^k = \sum_{k=0}^n \frac{(n+k)!}{k!} {n \choose k} x^k $$
And we can equate these and simplify with:
$$\sum_{k=0}^n \frac{(n+k)!}{k!} {n \choose k} x^k = \sum_{k=0}^n {n \choose k}^2 n! x^{n-k} (1+x)^{k} \\ \sum_{k=0}^n \frac{(n+k)!}{k!n!} {n \choose k} x^k = \sum_{k=0}^n {n \choose k}^2 x^{n-k} (1+x)^{k} $$
we would very specifically like $x^n$ term only and since $x^n$ only appear in the first term of the binomial expansion (if we expand $(1+x)^k x^{n-k}$ the coefficient of $x^n$ is always the first term) the coefficient is ${n \choose 0}$ so we obtain: $$\frac{2n!}{n!n!} {n \choose n} = \sum_{k=0}^n {n \choose k}^2 { n \choose 0} \\ {2n \choose n} = \sum_{k=0}^n {n \choose k}^2 $$ Q.E.D.
P.s. I read the title and saw analysis and wrote a proof before I noticed it said with a derivative so I wrote up this other proof and deleting it would make me sad. Here it is with a cauchy product.
The first trick is figuring out how to express $\sum {n \choose k}^2$ as a function, it is supposed to be $f(x) = (1+x)^n(1+x)^n = (1+x)^{2n}$ and you can see the equivalency via a cauchy product: $$ \begin{align}f(x) &= \left(\sum_{k=0}^n x^k {n \choose k} \right) \left(\sum_{k=0}^n x^k {n \choose k} \right) \\& = \sum_{r=0}^{2n} \left( \sum_{k=0}^r {n \choose k}{n \choose n-k}\right) x^r \\ &=\sum_{r=0}^{2n} \left( \sum_{k=0}^r {n \choose k}^2\right) x^r \end{align} $$
Then, it follows that since this is the binomial expansion for $2n$, at $r=n$, $\sum {n \choose k}^2 = {2n \choose n}$. Also note here that in the third step we use the fact that ${n \choose k} = {n \choose n -k}$.