Leibniz formula for the nth derivative of $f(x)=x^{n-1} \ln x$ Problem : Calculate the derivative of the function $f: ]0,+\infty\left[\longrightarrow \mathbb{R}\right.$ defined by $f(x)=x^{n-1} \ln x$.
Solution
Let $g_{1}(x)=x^{n-1}$ et $g_{2}(x)=\ln x .$ So we have$f=g_{1} g_{2}$
for all $k$ verifying $1 \leq k \leq n-1$ we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$
The Leibniz formula gives
$\begin{array}{c}
f^{(n)}(x)=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) g_{1}^{(k)}(x) g_{2}^{(n-k)}(x) \\
=x^{n-1} \frac{(-1)^{n-1}(n-1) !}{x^{n}}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) \cdots(n-1-k+1) x^{n-1-k} \frac{(-1)^{n-k-1}(n-k-1) !}{x^{n-k}} \\
=\frac{(-1)^{n-1}(n-1) !}{x}+\sum_{k=1}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(-1)^{n-k-1}(n-1) !}{x}\\
=-\frac{(n-1) !}{x} \cdot \sum_{k=0}^{n-1}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}
\end{array}$
The sum above is the same as $\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$ (prived of $k=n,$) which is equal to -1
We obtain
$$
f^{(n)}(x)=\frac{(n-1) !}{x}
$$
What I don't get:
What's the point of this step:
we have
$$
g_{1}^{(k)}(x)=(n-1) \cdots(n-1-k+1) x^{n-1-k}
$$
et
$$
g_{1}^{(n)}(x)=0
$$
and which solution is wrong?
I spent a lot of time already on this, I think it's a matter of indices I can't point, both solutions seem correct to me.
How I calculated :
we have that
$\ln ^{(n)}(x)=\frac{(-1)^{n-1}(n-1) !}{x^{n}}$
and
$(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$
\begin{aligned}
f^{(n)} &=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \\
&=\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right) \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! } \frac{(n-1-k) !(-1)^{n-1-k}}{x^{n-k}} \\
=& x^{-1} \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(n-1) !(-1)^{n-k-1}
\end{aligned}
$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k}$
which should give me zero? since
$\sum_{k=0}^{n}\left(\begin{array}{l}
n \\
k
\end{array}\right)(-1)^{n-k} = (1+(-1))^{n}=0 $
 A: I know I'm not answering the question, but I can't resist the temptation to show you a trick to arrive at the solution faster (no Leibniz formula): Integrate once before differentiating.
Let $f_n(x)=x^{n-1}\ln x$. Compute one of its primitives (antiderivatives) by integrating by parts:
$$\int_0^xf_n(t)dt=\frac{x^n}n \ln x-\frac{x^{n-1}}{n^2}=\frac{f_{n+1}(x)}n-\frac{x^{n-1}}{n^2}$$
So to obtain $f_n^{(n)}(x)$, we just need to differentiate the above $n+1$ times (noting that this will cancel the second term as it's a polynomial of degree $n-1$):
$$f_n^{(n)}(x)=\frac{f_{n+1}^{(n+1)}(x)}n$$
Solving this recurrence formula gives you the result $$f_n^{(n)}(x)=(n-1)!f_1^\prime(x)=\frac{(n-1)!}{x}$$
A: Your solution is almost correct (and neither $g_1^{(k)}(x)$ or $g_1^{(n)}(x)$ are wrong). The only problem is that $(x^{n-1})^{(k)} = \frac{(n-1) !x^{n-1-k}}{(n-1-k) ! }$ is only true for $0 \le k \le n-1$. For $k \ge n$, the $k$th derivative will be $0$. For example, the $2$nd derivative of $x^1$ is $0$.
The step
$$f^{(n)} =\sum_{k=0}^{n}\binom{n}{k}\left(x^{n-1}\right)^{(k)}(\ln x)^{(n-k)} \tag 1$$
is right. But then at the next step, the sum should only go from $k = 0$ to $k = n-1$ since $\left(x^{n-1}\right)^{(n)} = 0$.
Then after that, the same logic applies, so you would end up with $$= (-1)x^{-1} (n-1) ! \sum_{k=0}^{n-1}\binom{n}{k}(-1)^{n-k} = \frac{(n-1)!}{x}$$
