How to prove that $\frac{d^n}{dx^n}(x^2-1)^n=0$ has $n$ real roots? How do I prove that $$\frac{d^n}{dx^n}(x^2-1)^n=0$$ has $n$ real roots?
 A: Firstly, $\partial(\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n])=n$, so $\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n]$ has at most $n$ roots.
Applying Leibniz Identity,
$$\forall k<n,k\in\mathbb{N},\quad \dfrac{\text{d}^k}{\text{d}x^k}[(x^2-1)^n]=\sum\limits_{i=0}^k\binom{k}{i}\dfrac{\text{d}^i(x-1)^n}{\text{d}x^i}\dfrac{\text{d}^{k-1}(x+1)^n}{\text{d}x^{k-i}}$$
Thus, $\pm1$ are roots of $\dfrac{\text{d}^k}{\text{d}x^k}[(x^2-1)^n]$ when $k<n$.
Applying Rolle's Theorem, $\dfrac{\text{d}}{\text{d}x}[(x^2-1)^n]$ has roots $\xi_{11}\in(-1,1)$ and $\pm1$.
Applying Rolle's Theorem again, $\dfrac{\text{d}^2}{\text{d}x^2}[(x^2-1)^n]$ has roots $\xi_{21}\in(-1,\xi_{11})$, $\xi_{22}\in(\xi_{11},1)$ and $\pm1$.
Similarly, $\dfrac{\text{d}^{n-1}}{\text{d}x^{n-1}}[(x^2-1)^n]$ has $n+1$ roots
$$-1=\xi_{n-1,0}<\xi_{n-1,1}<\dots<\xi_{n-1,n-1}<\xi_{n-1,n}=1$$
At last, Applying Rolle's Theorem again, we get the $n$ roots of $\dfrac{\text{d}^n}{\text{d}x^n}[(x^2-1)^n]$:
$$\xi_{n,1}\in(-1,\xi_{n-1,1}),\dots,\xi_{n,n}\in(\xi_{n-1,n-1},1)$$
Q.E.D.
A: Hint: Try Induction on $n$. 
Also on the above remark, between any two real roots of $f$, there will be a real root of $f'$.
A: For those who are interested, there is a more general result one can use:
For any polynomial $f(z)$ with complex coefficients, the roots
of $f'(z)$ is contained within the convex hull of the roots of $f(z)$.
Since $p(x) = (x^2-1)^n$ has only 2 distinct roots $\pm 1$,
all roots of $p'(x)$ belongs to the line segment joining $\pm 1$.
Namely, the interval $[-1,1]$ on the real axis. Apply this theorem $n$ times,
we find all roots of $\frac{d^n}{dx^n} p(x)$ are not only real but belong to
$[-1,1]$.
The proof of the general case is pretty short, let I give it here.
Let $f(z) = C \prod_i (z - \alpha_i)^{m_i}$ be a complex polynomial with roots $\alpha_i$, each with multiplicity $m_i$.
Let $w$ be a root of $f'(z)$. If $w$ is one of $\alpha_i$, then we are done. If not, we have:
$$0 = f'(w)/f(w) = \sum_i \frac{m_i}{w - \alpha_i}
    = \sum_i \frac{m_i (\bar{w} - \bar{\alpha_i})}{|w - \alpha_i|^2}$$
Taking complex conjugate and rearrange terms, we get
$w = \frac{\sum_i \rho_i \alpha_i}{\sum_i \rho_i}$
where $\rho_i = \frac{m_i}{|w - \alpha_i|^2}$ are all real and positive.
A: Let $P(x)$ be a polynomial of degree $n$. Suppose $P(x)$ has a zero of degree $m$ at $x=a$. This means that
$$
P(x)=(x-a)^mQ(x)
$$
where $Q(x)$ is a polynomial of degree $n-m$ and $Q(a)\ne0$. Thus,
$$
\begin{align}
P'(x)
&=(x-a)^{m-1}\left[mQ(x)+(x-a)Q'(x)\right]\\
&=(x-a)^{m-1}R(x)
\end{align}
$$
where $R(x)$ is degree $n-m$ and $R(a)=mQ(a)\ne0$.
That is, each zero of $P(x)$ is reduced by one degree. However, Rolle's Theorem insures that between two adjacent zeros of $P(x)$, $P'(x)$ has a zero.
Let $k$ be the number of zeros of $P(x)$ not counting multiplicities. Differentiation will reduce the degree of each of the $k$ zeros by one, and will add $k-1$ zeros between the $k$ zeros by Rolle's Theorem. Thus, the total number of zeros of $P'(x)$, counting multiplicities, is at least the number of zeros of $P(x)$, counting multiplicities, minus one. New zeros might be introduced to $P'(x)$.
The number of zeros not counting multiplicities of $P'(x)$ is at least $k-1$. In addition, each multiple zero of $P(x)$ persists. Thus, $P'(x)$ has at least $k+m-1$ roots not counting multiplicities where $P(x)$ has $k$ zeros, not counting multiplicities including $m$ multiple zeros.
Counting multiplicities, $(x^2-1)^n$ has $2$ zeros of degree $n$. Therefore, these two roots persist until the $n^{\text{th}}$ derivtive. After $n-1$ derivatives, there should be at least $n+1$ roots with the multiple roots gone. After the $n^{\text{th}}$ derivative, there should be at least $n$ roots, not counting multiplicity. As a degree $n$ polynomial, it has at most $n$ roots. Therefore, it must have exactly $n$ roots.
