Here is a different way to prove the same result as @LutzL:
Theorem 1. Let $n\in\mathbb{N}$. Then, the derivative of the polynomial
$\dbinom{x}{n}\in\mathbb{Q}\left[ x\right] $ is
\begin{equation}
\dfrac{d}{dx}\dbinom{x}{n}=\sum_{k=1}^{n}\dfrac{\left( -1\right) ^{k-1}}
{k}\dbinom{x}{n-k}.
\end{equation}
My proof relies on three identities for binomial coefficients. The first is
the famous Chu-Vandermonde identity:
Theorem 2. Let $n\in\mathbb{N}$. Then, in the polynomial ring
$\mathbb{Q}\left[ x,y\right] $ (in two indeterminates $x$ and $y$), we have
\begin{equation}
\dbinom{x+y}{n}=\sum_{k=0}^{n}\dbinom{x}{k}\dbinom{y}{n-k}.
\end{equation}
For a direct proof of Theorem 2, see Theorem 3.30 in Darij Grinberg, Notes
on the combinatorial fundamentals of algebra, 10 January
2019 (where I
denote the indeterminates $x$ and $y$ by $X$ and $Y$ to visually keep them
apart from two numbers $x$ and $y$). That said, if you know any proof of the
classical Vandermonde convolution identity
\begin{equation}
\dbinom{a+b}{n}=\sum_{k=0}^{n}\dbinom{a}{k}\dbinom{b}{n-k}\qquad\text{for
}a\in\mathbb{N}\text{ and }b\in\mathbb{N}
\end{equation}
(see also the Cut-the-Knot
page),
then you can easily leverage it to obtain a proof of Theorem 2 by applying the
standard polynomial identity trick (if two univariate polynomials agree on
$\mathbb{N}$, then they are identical) twice (since there are two indeterminates).
The next identity I need is a simple one (one of the forms of the absorption identity):
Proposition 3. Let $k$ be a positive integer. Then, in the polynomial ring
$\mathbb{Q}\left[ y\right] $ (in one indeterminate $y$), we have
\begin{equation}
\dbinom{y}{k}=\dfrac{y}{k}\dbinom{y-1}{k-1}.
\end{equation}
Proof of Proposition 3. The definition of $\dbinom{y-1}{k-1}$ yields
\begin{align*}
\dbinom{y-1}{k-1} & =\dfrac{\left( y-1\right) \left( y-2\right)
\cdots\left( \left( y-1\right) -\left( k-1\right) +1\right) }{\left(
k-1\right) !}\\
& =\dfrac{\left( y-1\right) \left( y-2\right) \cdots\left( y-k+1\right)
}{\left( k-1\right) !}.
\end{align*}
Multiplying this equality by $\dfrac{y}{k}$, we find
\begin{align*}
\dfrac{y}{k}\dbinom{y-1}{k-1} & =\dfrac{y}{k}\cdot\dfrac{\left( y-1\right)
\left( y-2\right) \cdots\left( y-k+1\right) }{\left( k-1\right)
!}=\dfrac{y\cdot\left( y-1\right) \left( y-2\right) \cdots\left(
y-k+1\right) }{k\cdot\left( k-1\right) !}\\
& =\dfrac{y\left( y-1\right) \cdots\left( y-k+1\right) }{k!}
\end{align*}
(since $y\cdot\left( y-1\right) \left( y-2\right) \cdots\left(
y-k+1\right) =y\left( y-1\right) \cdots\left( y-k+1\right) $ and
$k\cdot\left( k-1\right) !=k!$). Comparing this with
\begin{equation}
\dbinom{y}{k}=\dfrac{y\left( y-1\right) \cdots\left( y-k+1\right) }
{k!}\qquad\left( \text{by the definition of }\dbinom{y}{k}\right) ,
\end{equation}
we obtain $\dbinom{y}{k}=\dfrac{y}{k}\dbinom{y-1}{k-1}$. This proves
Proposition 3. $\blacksquare$
The final identity I need is so simple even I am leaving it as an exercise:
Proposition 4. Let $m\in\mathbb{N}$. Then, $\dbinom{-1}{m}=\left(
-1\right) ^{m}$.
(Just in case: Proposition 4 is Corollary 3.17 in
Darij Grinberg, Notes on the combinatorial fundamentals of algebra,
10 January 2019.)
Now, we define a weird notation:
Definition. Let $R\in\mathbb{Q}\left[ x,y\right] $ be a polynomial in
two indeterminates $x$ and $y$. Let $a\in\mathbb{Q}$. Then,
$\operatorname*{ev}\limits_{y\to a}R$ shall denote the result of
substituting $a$ for $y$ in $R$. This is a polynomial in $\mathbb{Q}\left[
x\right] $.
For example, $\operatorname*{ev}\limits_{y\to 3}\left( x+y\right)
^{2}=\left( x+3\right) ^{2}$.
For the algebraists in the room: If we fix $a\in\mathbb{Q}$, then the map
$\operatorname*{ev}\limits_{y\to a}$ (sending each $R\in
\mathbb{Q}\left[ x,y\right] $ to $\operatorname*{ev}\limits_{y\to
a}R\in\mathbb{Q}\left[ x\right] $) is a $\mathbb{Q}\left[ x\right]
$-algebra homomorphism, called an evaluation homomorphism. This is why I am
using "ev" as its name. The symbol "$\operatorname*{ev}\limits_{y\to
a}$" should you remind you of "$\lim\limits_{y\to a}$" from calculus,
and that similarity is substantial: For example,
\begin{equation}
\operatorname*{ev}\limits_{y\to 1}\underbrace{\dfrac{y^{3}-1}{y-1}
}_{=y^{2}+y+1}=\operatorname*{ev}\limits_{y\to 1}\left( y^{2}
+y+1\right) =1^{2}+1+1=3,
\end{equation}
although you cannot simply substitute $1$ into the expression
$\dfrac{y^{3}-1}{y-1}$. So,
when $R$ is a fraction of two polynomials that happens to itself be a
polynomial, then $\operatorname*{ev}\limits_{y\to a}R$ really is
something like "the limit of $R$ as $y$ approaches $a$", because in order to
compute $\operatorname*{ev}\limits_{y\to a}R$, we must first rewrite
the fraction as an actual polynomial and only then substitute $a$ for $y$.
Finally, we shall need the following formula for the derivative of a polynomial,
which is the algebraic counterpart to the classical analytical definition of
the derivative:
Theorem 5. Let $P\in\mathbb{Q}\left[ x\right] $ be a polynomial.
Consider the polynomial $P\left( x+y\right) \in\mathbb{Q}\left[ x,y\right]
$ in two indeterminates $x$ and $y$. Then, the polynomial $P\left(
x+y\right) -P\left( x\right) \in\mathbb{Q}\left[ x,y\right] $ is
divisible by $y$, and we have
\begin{equation}
\dfrac{d}{dx}P=\operatorname*{ev}\limits_{y\to 0}\dfrac{P\left(
x+y\right) -P\left( x\right) }{y}.
\end{equation}
Proof of Theorem 5. Write the polynomial $P$ in the form
$P = \sum\limits_{i=0}^{n}a_{i}x^{i}$ for some $n\in\mathbb{N}$
and $a_{0},a_{1},\ldots,a_{n} \in\mathbb{Q}$.
We WLOG assume that $n\geq 1$ (since otherwise, we can increase
$n$ by adding zero coefficients to $P$).
From $P = \sum\limits_{i=0}^{n}a_{i}x^{i}$, we obtain
\begin{equation}
\dfrac{d}{dx}P=\sum_{i=1}^{n}a_{i}ix^{i-1}
\label{darij.pf.t4.der}
\tag{1}
\end{equation}
(by the definition of $\dfrac{d}{dx}P$). On the other hand, from
$P = \sum\limits_{i=0}^{n}a_{i}x^{i}$, we obtain
\begin{align*}
P\left( x+y\right) & =\sum_{i=0}^{n}a_{i}\underbrace{\left( x+y\right)
^{i}}_{\substack{=\left( y+x\right) ^{i}\\=
\sum\limits_{j=0}^{i}\dbinom{i}{j}
y^{j}x^{i-j}\\\text{(by the binomial formula)}}}
=\sum_{i=0}^{n}a_{i}
\underbrace{\sum\limits_{j=0}^{i}\dbinom{i}{j}y^{j}x^{i-j}
}_{\substack{=\dbinom{i}{0}y^{0}x^{i-0}
+\sum\limits_{j=1}^{i}\dbinom{i}{j}y^{j}x^{i-j}\\\text{(here, we have
split off the}\\\text{addend for }j=0\text{ from the sum)}}}\\
& =\sum_{i=0}^{n}a_{i}\left( \underbrace{\dbinom{i}{0}}_{=1}
\underbrace{y^{0}}_{=1}\underbrace{x^{i-0}}_{=x^{i}}+\sum_{j=1}^{i}\dbinom
{i}{j}y^{j}x^{i-j}\right) \\
& =\sum_{i=0}^{n}a_{i}\left( x^{i}+\sum_{j=1}^{i}\dbinom{i}{j}y^{j}
x^{i-j}\right) =\underbrace{\sum_{i=0}^{n}a_{i}x^{i}}_{=P=P\left( x\right)
}+\sum_{i=0}^{n}a_{i}\sum_{j=1}^{i}\dbinom{i}{j}y^{j}x^{i-j}\\
& =P\left( x\right) +\sum_{i=0}^{n}a_{i}\sum_{j=1}^{i}\dbinom{i}{j}
y^{j}x^{i-j}.
\end{align*}
Subtracting $P\left( x\right) $ from this equality, we find
\begin{align}
& P\left( x+y\right) -P\left( x\right) \nonumber\\
& =\sum_{i=0}^{n}a_{i}\sum_{j=1}^{i}\dbinom{i}{j}y^{j}x^{i-j}
=\underbrace{\sum\limits_{i=0}^{n}\sum\limits_{j=1}^{i}
}_{=\sum\limits_{j=1}^{n}\sum\limits_{i=j}^{n}
}a_{i}\dbinom{i}{j}y^{j}x^{i-j}=\sum_{j=1}^{n}\sum_{i=j}^{n}a_{i}\dbinom{i}
{j}y^{j}x^{i-j}\nonumber\\
& =\sum_{j=1}^{n}\underbrace{y^{j}}_{\substack{=yy^{j-1}\\\text{(since }
j\geq1\text{)}}}\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}=\sum_{j=1}^{n}
yy^{j-1}\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}\nonumber\\
& =y\sum_{j=1}^{n}y^{j-1}\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}
.
\label{darij.pf.t4.1}
\tag{2}
\end{align}
Hence, this polynomial $P\left( x+y\right) -P\left( x\right) $ is
divisible by $y$. Moreover, dividing the equality \eqref{darij.pf.t4.1} by
$y$, we obtain
\begin{equation}
\dfrac{P\left( x+y\right) -P\left( x\right) }{y}=\sum_{j=1}^{n}y^{j-1}
\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}.
\end{equation}
Hence,
\begin{align*}
\operatorname*{ev}\limits_{y\to 0}\dfrac{P\left( x+y\right) -P\left(
x\right) }{y} & =\operatorname*{ev}\limits_{y\to 0}\sum_{j=1}
^{n}y^{j-1}\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}=\sum_{j=1}^{n}0^{j-1}
\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}\\
& \qquad\left(
\begin{array}
[c]{c}
\text{by the definition of }\operatorname*{ev}\limits_{y\to 0}\text{,
because the }y^{j-1}\\
\text{are genuine polynomials in }y
\end{array}
\right) \\
& =\underbrace{0^{1-1}}_{=1}\sum_{i=1}^{n}a_{i}\underbrace{\dbinom{i}{1}}
_{=i}x^{i-1}+\sum_{j=2}^{n}\underbrace{0^{j-1}}_{\substack{=0\\\text{(since
}j-1>0\\\text{(because }j\geq2\text{))}}}\sum_{i=j}^{n}a_{i}\dbinom{i}
{j}x^{i-j}\\
& \qquad\left( \text{here, we have split off the addend for }j=1\text{ from
the sum}\right) \\
& =\underbrace{\sum_{i=1}^{n}a_{i}ix^{i-1}}_{\substack{=\dfrac{d}
{dx}P\\\text{(by \eqref{darij.pf.t4.der})}}}+\underbrace{\sum_{j=2}^{n}
0\sum_{i=j}^{n}a_{i}\dbinom{i}{j}x^{i-j}}_{=0}=\dfrac{d}{dx}P.
\end{align*}
In other words, $\dfrac{d}{dx}P=\operatorname*{ev}\limits_{y\to
0}\dfrac{P\left( x+y\right) -P\left( x\right) }{y}$. This completes the
proof of Theorem 5. $\blacksquare$
Now, we can prove Theorem 1:
Proof of Theorem 1. Theorem 5 (applied to $P=\dbinom{x}{n}$)
shows that the polynomial $\dbinom{x+y}{n}-\dbinom{x}{n}\in\mathbb{Q}
\left[ x,y\right] $ is divisible by $y$, and that we have
\begin{equation}
\dfrac{d}{dx}\dbinom{x}{n}=\operatorname*{ev}\limits_{y\to 0}
\dfrac{\dbinom{x+y}{n}-\dbinom{x}{n}}{y}.
\label{darij.pf.t1.1}
\tag{3}
\end{equation}
Now,
\begin{align*}
\dbinom{x+y}{n} & =\dbinom{y+x}{n}=\sum_{k=0}^{n}\dbinom{y}{k}\dbinom{x}
{n-k}\qquad\left(
\begin{array}
[c]{c}
\text{by Theorem 2, applied to }y\text{ and }x\\
\text{instead of }x\text{ and }y
\end{array}
\right) \\
& =\underbrace{\dbinom{y}{0}}_{=1}\underbrace{\dbinom{x}{n-0}}_{=\dbinom{x}
{n}}+\sum_{k=1}^{n}\dbinom{y}{k}\dbinom{x}{n-k}\\
& \qquad\left( \text{here, we have split off the addend for }k=0\text{ from
the sum}\right) \\
& =\dbinom{x}{n}+\sum_{k=1}^{n}\dbinom{y}{k}\dbinom{x}{n-k}.
\end{align*}
Subtracting $\dbinom{x}{n}$ from this equality, we find
\begin{align*}
\dbinom{x+y}{n}-\dbinom{x}{n} & =\sum_{k=1}^{n}\underbrace{\dbinom{y}{k}
}_{\substack{=\dfrac{y}{k}\dbinom{y-1}{k-1}\\\text{(by Proposition 3)}
}}\dbinom{x}{n-k}\\
& =\sum_{k=1}^{n}\dfrac{y}{k}\dbinom{y-1}{k-1}\dbinom{x}{n-k}=y\sum_{k=1}
^{n}\dfrac{1}{k}\dbinom{y-1}{k-1}\dbinom{x}{n-k}.
\end{align*}
Dividing this equality by $y$, we find
\begin{equation}
\dfrac{\dbinom{x+y}{n}-\dbinom{x}{n}}{y}=\sum_{k=1}^{n}\dfrac{1}{k}
\dbinom{y-1}{k-1}\dbinom{x}{n-k}.
\end{equation}
Now, \eqref{darij.pf.t1.1} becomes
\begin{align*}
\dfrac{d}{dx}\dbinom{x}{n} & =\operatorname*{ev}\limits_{y\to
0}\underbrace{\dfrac{\dbinom{x+y}{n}-\dbinom{x}{n}}{y}}_{
=\sum\limits_{k=1}^{n}
\dfrac{1}{k}\dbinom{y-1}{k-1}\dbinom{x}{n-k}}\\
& =\operatorname*{ev}\limits_{y\to 0}\sum_{k=1}^{n}\dfrac{1}{k}
\dbinom{y-1}{k-1}\dbinom{x}{n-k} \\
& =\sum_{k=1}^{n}\dfrac{1}{k}\underbrace{\dbinom
{0-1}{k-1}}_{\substack{=\dbinom{-1}{k-1}=\left( -1\right) ^{k-1}\\\text{(by
Proposition 4, applied to }m=k-1\\\text{(since }k-1\in\mathbb{N}\text{))}
}}\dbinom{x}{n-k}\\
& \qquad \left(\begin{array}{c}
\text{by the definition of } \operatorname*{ev}\limits_{y\to 0}\text{,} \\
\text{since the } \dbinom{y-1}{k-1} \text{ are genuine polynomials in } y
\end{array} \right) \\
& =\sum_{k=1}^{n}\dfrac{1}{k}\left( -1\right) ^{k-1}\dbinom{x}{n-k}
=\sum_{k=1}^{n}\dfrac{\left( -1\right) ^{k-1}}{k}\dbinom{x}{n-k}.
\end{align*}
This proves Theorem 1. $\blacksquare$