Determine: 1° (d^n(x²-1)^n)/(dx^n); 2° its value at x = 1
Determine: 1° $\dfrac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}}$; 2° its value at $x = 1$ and $x = -1$; 3° $\dfrac{d^{\,n-1}\bigl(x^{2}-1\bigr)^{n}}{dx^{\,n-1}}$ at $x = 1$ and $x = -1$.
Explanation of Method
Factoring first makes the endpoint values fall out of a single Leibniz term.
Step 1: Factor, then apply Leibniz
Writing $\bigl(x^{2}-1\bigr)^{n} = \bigl(x-1\bigr)^{n}\bigl(x+1\bigr)^{n}$ turns the problem into a product of two simple powers, whose derivatives are
\[ \Bigl[\bigl(x-1\bigr)^{n}\Bigr]^{(k)} = \frac{n!}{\bigl(n-k\bigr)!}\bigl(x-1\bigr)^{\,n-k}, \qquad \Bigl[\bigl(x+1\bigr)^{n}\Bigr]^{(m-k)} = \frac{n!}{\bigl(n-m+k\bigr)!}\bigl(x+1\bigr)^{\,n-m+k}. \tag{L} \]
Step 2: Part 1°: the $n$-th derivative
Take $m = n$ in (L), so that the second factor becomes $\dfrac{n!}{k!}\bigl(x+1\bigr)^{k}$: \[\frac{d^{n}}{dx^{n}}\bigl(x^{2}-1\bigr)^{n} = \sum_{k=0}^{n}\binom{n}{k}\frac{n!}{\bigl(n-k\bigr)!}\cdot\frac{n!}{k!}\,\bigl(x-1\bigr)^{\,n-k}\bigl(x+1\bigr)^{k}. \tag{1}\]
\[\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}} = \sum_{k=0}^{n}\binom{n}{k}\frac{n!}{\bigl(n-k\bigr)!}\cdot\frac{n!}{k!}\bigl(x-1\bigr)^{\,n-k}\bigl(x+1\bigr)^{k}\]
Step 3: Part 2°: value at $x = 1$
Every term of (1) carries the factor $\bigl(x-1\bigr)^{\,n-k}$, which vanishes at $x = 1$ unless $k = n$. Only that one term survives: \[\binom{n}{n}\cdot\frac{n!}{0!}\cdot\frac{n!}{n!}\cdot\bigl(1-1\bigr)^{0}\bigl(1+1\bigr)^{n} = n!\cdot2^{n}.\]
\[\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}}\bigg|_{x=1} = 2^{n}\,n!\]
Step 4: Part 2°: value at $x = -1$
Now it is $\bigl(x+1\bigr)^{k}$ that vanishes, unless $k = 0$: \[\binom{n}{0}\cdot\frac{n!}{n!}\cdot\frac{n!}{0!}\cdot\bigl(-1-1\bigr)^{n}\bigl(-1+1\bigr)^{0} = n!\,\bigl(-2\bigr)^{n}.\]
\[\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}}\bigg|_{x=-1} = \bigl(-2\bigr)^{n}n! = \bigl(-1\bigr)^{n}2^{n}\,n!\]
Step 5: Part 3°: one derivative fewer
Take $m = n-1$ in (L); the second factor is then $\dfrac{n!}{\bigl(k+1\bigr)!}\bigl(x+1\bigr)^{k+1}$: \[\frac{d^{\,n-1}}{dx^{\,n-1}}\bigl(x^{2}-1\bigr)^{n} = \sum_{k=0}^{n-1}\binom{n-1}{k}\frac{n!}{\bigl(n-k\bigr)!}\cdot\frac{n!}{\bigl(k+1\bigr)!}\bigl(x-1\bigr)^{\,n-k}\bigl(x+1\bigr)^{\,k+1}. \tag{2}\]
Now both exponents are strictly positive for every $k$ in range: $n-k\ge1$ since $k\le n-1$, and $k+1\ge1$ always. So every term vanishes at either endpoint: \[\frac{d^{\,n-1}\bigl(x^{2}-1\bigr)^{n}}{dx^{\,n-1}}\bigg|_{x=1} = 0, \qquad \frac{d^{\,n-1}\bigl(x^{2}-1\bigr)^{n}}{dx^{\,n-1}}\bigg|_{x=-1} = 0\]
Step 6: Verification
\[n = 2:\ \bigl(x^{2}-1\bigr)^{2} = x^{4}-2x^{2}+1,\quad \frac{d^{2}}{dx^{2}} = 12x^{2}-4\]
\[\text{At } x = \pm1:\ 12-4 = 8 = 2^{2}\cdot2! \;\;\checkmark; \qquad \frac{d}{dx} = 4x^{3}-4x = 0 \text{ at } x = \pm1 \;\;\checkmark\]
\[n = 3:\ \frac{d^{3}}{dx^{3}}\bigl(x^{2}-1\bigr)^{3} = 120x^{3}-72x;\ \text{at } 1:\ 48 = 2^{3}3!;\ \text{at } -1:\ -48 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}} = \sum_{k=0}^{n}\binom{n}{k}\frac{n!}{\bigl(n-k\bigr)!}\cdot\frac{n!}{k!}\bigl(x-1\bigr)^{\,n-k}\bigl(x+1\bigr)^{k}\,}\]
Final answer:
\[\boxed{\,\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}}\bigg|_{x=1} = 2^{n}\,n!\,}\]
Final answer:
\[\boxed{\,\frac{d^{n}\bigl(x^{2}-1\bigr)^{n}}{dx^{n}}\bigg|_{x=-1} = \bigl(-2\bigr)^{n}n! = \bigl(-1\bigr)^{n}2^{n}\,n!\,}\]
Final answer:
\[\boxed{\,\frac{d^{\,n-1}\bigl(x^{2}-1\bigr)^{n}}{dx^{\,n-1}}\bigg|_{x=1} = 0, \qquad \frac{d^{\,n-1}\bigl(x^{2}-1\bigr)^{n}}{dx^{\,n-1}}\bigg|_{x=-1} = 0\,}\]
Remark
Remark. By Rodrigues' formula the expression in part 1° equals $2^{n}n!\,P_{n}(x)$, where $P_{n}$ is the $n$-th Legendre polynomial; parts 2° and 3° are then the familiar normalisations $P_{n}(1) = 1$ and $P_{n}(-1) = \bigl(-1\bigr)^{n}$. Part 3° is what makes the repeated integration by parts in the Legendre orthogonality proof work: all the boundary terms vanish.