High-order derivative identity

Official Solution :

Of course, you could prove it by induction, but such proof would be limited and not elegant. I will prove it directly:

We know that we can solve the high-order derivative by using Leibniz's formula, which is a calculus analog of the binomial theorem:

$\displaystyle (g(x)h(x))^{(n)}= \sum_{k=0}^{n} {n \choose k}g^{(n-k)}(x)h^{(k)}(x)$

Let $g^{(0)}(x)=\sin x$ and $h^{(0)}(x)=\dfrac{1}{x}$ .

Notice that $\displaystyle h^{(0)}(x)=x^{-1},\ h^{(1)}(x)=(-1)x^{-2},\ h^{(2)}(x)=(-1)(-2)x^{-3},\ h^{(3)}(x)=(-1)(-2)(-3)x^{-4},...$

So that $\displaystyle h^{(n)}(x)=(-1)^n n! x^{-n-1}$

$\displaystyle nf^{(n-1)}(x)=n {n-1 \choose 0}g^{(n-1)}(x)h^{(0)}(x) + n {n-1 \choose 1}g^{(n-2)}(x)h^{(1)}(x)+n {n-1 \choose 2}g^{(n-3)}(x)h^{(2)}(x)+......+n {n-1 \choose n-1}g^{(0)}(x)h^{(n-1)}(x) \\ =n\dfrac{(n-1)!}{0!(n-1)!}g^{(n-1)}(x)(-1)^{0}0!x^{-1}+n\dfrac{(n-1)!}{1!(n-2)!}g^{(n-2)}(x)(-1)^{1}1!x^{-2}+n\dfrac{(n-1)!}{2!(n-3)!}g^{(n-3)}(x)(-1)^{2}2!x^{-3}+...... \\+n\dfrac{(n-1)!}{(n-1)!0!}g^{0}(x)(-1)^{n-1}(n-1)!x^{-n} \\ =\boxed {\dfrac{n!}{(n-1)!}g^{(n-1)}(x)(-1)^{0}x^{-1}+\dfrac{n!}{(n-2)!}g^{(n-2)}(x)(-1)^{1}x^{-2}+\dfrac{n!}{(n-3)!}g^{(n-3)}(x)(-1)^{2}x^{-3}+......+\dfrac{n!}{0!}g^{0}(x)(-1)^{n-1}x^{-n}}$

$\displaystyle xf^{(n)}(x)=x {n \choose 0}g^{(n)}(x)h^{(0)}(x) + x {n \choose 1}g^{(n-1)}(x)h^{(1)}(x)+x {n \choose 2}g^{(n-2)}(x)h^{(2)}(x)+......+x {n \choose n}g^{(0)}(x)h^{(n)}(x) \\ =x\dfrac{n!}{0!n!}g^{(n)}(x)(-1)^{0}0!x^{-1}+x\dfrac{n!}{1!(n-1)!}g^{(n-1)}(x)(-1)^{1}1!x^{-2}+x\dfrac{n!}{2!(n-2)!}g^{(n-2)}(x)(-1)^{2}2!x^{-3}+......+x\dfrac{n!}{n!0!}g^{0}(x)(-1)^{n}(n)!x^{-n-1} \\ =\dfrac{n!}{0!n!}g^{(n)}(x)(-1)^{0}0!x^{0}+\dfrac{n!}{1!(n-1)!}g^{(n-1)}(x)(-1)^{1}1!x^{-1}+\dfrac{n!}{2!(n-2)!}g^{(n-2)}(x)(-1)^{2}2!x^{-2}+......+\dfrac{n!}{n!0!}g^{0}(x)(-1)^{n}n!x^{-n} \\ =\boxed{\dfrac{n!}{n!}g^{(n)}(x)(-1)^{0}x^{0}+\dfrac{n!}{(n-1)!}g^{(n-1)}(x)(-1)^{1}x^{-1}+\dfrac{n!}{(n-2)!}g^{(n-2)}(x)(-1)^{2}x^{-2}+......+\dfrac{n!}{0!}g^{0}(x)(-1)^{n}x^{-n}}$

Notice that $xf^{(n)}(x)$ has $n+1$ terms and $nf^{(n-1)}(x)$ has $n$ terms. From the second term of $xf^{(n)}(x)$ , when added up together, the terms cancel with each other since the negative sign alternates, left with the first term of $xf^{(n)}(x)$ , which simplifies to $g^{(n)}(x)$ .

So $|nf^{(n-1)}(x) + xf^{(n)}(x)|=|g^{(n)}(x)|$ .

Since $g^{(0)}(x)=\sin x,\ g^{(1)}(x)=\cos x,\ g^{(2)}(x)=-\sin x,\ g^{(3)}(x)=-\cos x,\ g^{(4)}(x)=\sin x$ , whose derivative has a period of 4, when $x=\dfrac{\pi}{4}$ , $|g^{(n)}(\dfrac{\pi}{4})|=\dfrac{\sqrt{2}}{2}$ .

We get $|nf^{(n-1)}(\dfrac{\pi}{4}) + xf^{(n)}(\dfrac{\pi}{4})|=|g^{(n)}(\dfrac{\pi}{4})|=\dfrac{\sqrt{2}}{2}=A$ .

The answer is $\lfloor 10000A \rfloor=\boxed {7071}$ .