$2n$th derivative of $(x^2-1)^n$

Question

How is the derivative $\frac{d^{2n}}{dx^{2n}}(x^2-1)^n=(2n)!$. I have tried to use Leibniz's formula but couldn't reach the solution.

Answer 1

The leading term of the polynomial is $x^{2n},$ so \begin{align} p'(x) &= 2nx^{2n-1}+\ldots,\\ p''(x) &= 2n(2n-1)x^{2n-2}+\ldots,\\ p'''(x) &= 2n(2n-1)(2n-2)x^{2n-3}+\ldots,\\ \ldots\\ p^{(2n-1)}(x) &= 2n(2n-1)(2n-2)\cdot\ldots\cdot3\cdot2x\\ p^{(2n)}(x) &= 2n(2n-1)(2n-2)\cdot\ldots\cdot3\cdot2\cdot1 \end{align} while the other terms disappear along the calculation.

Answer 2

Hint: you can show with induction that the $n$th derivative of a polynomial of $n$ degree is $a_n \cdot (n)!$

Answer 3

$$(x^{2}-1)^{n} = x^{2n}-{{n}\choose{1}} x^{2n-2} + {{n}\choose{2}} x^{2n-4} + \cdot \cdot \cdot + (-1)^{n} $$ $$\frac{d^{2n}}{(dx)^{2n}} x^{2n} = (2n)! $$ ,,, because for others term, derivative is $0$.

Answer 4

Another approach is to write $ \ f(x) \ = \ (x^2 - 1)^n \ = \ \underbrace{(x + 1)^n}_{g(x)}·\underbrace{(x - 1)^n}_{h(x)} \ \ , $ for which we have $ \ g^{(n)}(x) \ = \ h^{(n)}(x) \ = \ n! \ $ and $ \ g^{(k)}(x) \ = \ h^{(k)}(x) \ = \ 0 \ \ , $ for $ \ k \ > \ n \ \ . $ The "general Leibniz rule" then gives us $$ f^{(2n)} \ \ = \ \ \binom{2n}{0}·g^{(2n)}·h \ + \ \binom{2n}{1}·g^{(2n-1)}·h' \ + \ \binom{2n}{2}·g^{(2n-2)}·h'' $$ $$ + \ \ldots \ + \ \binom{2n}{n - 1}·g^{(n+1)}·h^{(n-1)} \ + \ \binom{2n}{n}·g^{(n)}·h^{(n)} \ + \ \binom{2n}{n + 1}·g^{(n-1)}·h^{(n+1)} $$ $$ + \ \ldots \ + \ \binom{2n}{2n - 2}·g''·h^{(2n-2)} \ + \ \binom{2n}{2n - 1}·g'·h^{(2n-1)} \ + \ \binom{2n}{2n}·g·h^{(2n)} \ \ . $$

Every product term in this "binomial" expansion is then equal to zero, except for the central one, which is $$ \binom{2n}{n} \ · \ g^{(n)}(x) \ · \ h^{(n)}(x) \ \ = \ \ \frac{(2n)!}{n! \ · \ n!} \ · \ n! \ · \ n! \ \ . $$ Hence, $ \ f^{(2n)}(x) \ \ = \ \ (2n)! \ \ . $