Fórmula de Leibniz de la derivada enésima

Demostramos la  fórmula de Leibniz de la derivada enésima y proprcionamos ejercicioss de aplicación.

Enunciado
1. Desarrollar la fórmula de Leibniz en el caso $n=4.$
2. Siendo $f(x)=\sqrt{x}\log (x+1)$ calcular $f^{(4)}(1)$ usando la fórmula de Leibniz.
3. Siendo $f(x)=e^x\operatorname{sen}x$ calcular $f^{(4)}(\pi/2).$
4. Usando la fórmula de Leibniz, hallar la derivada enésima de la función $f(x) = e^ {\alpha x} x^2.$
5. Demostrar la fórmula de Leibniz en los casos $n=2$ y $n=3.$
6. Demostrar la fórmula de Leibniz para la derivada enésima del producto:
   Sean $u,v:I\to \mathbb{R}$ dos funciones con derivadas hasta orden $n$ en todos los puntos del intervalo $I\subset \mathbb{R}.$ Entonces, se verifica en $I:$ $$(uv)^{(n)}=\sum_{k=0}^n\binom{n}{k}u^{(n-k)}v^{(k)},$$ en donde $u^{(0)}$ denota $u$ y $v^{(0)}$ denota $v.$
7. Usando la fórmula de Leibniz, calcular $y^{(n)}$ para $y=xe^x.$
8. Calcular $f^{(2016)}(0)$, siendo $f(x)=x^2\log (x+1)$.

Solución
1. $$(uv)^{(4)}=\displaystyle\binom{4}{0}u^{(4)}v^{(0)}+\displaystyle\binom{4}{1}u^{(3)}v^{(1)}+\displaystyle\binom{4}{2}u^{(2)}v^{(2)}+\displaystyle\binom{4}{3}u^{(1)}v^{(3)}$$ $$+\displaystyle\binom{4}{4}u^{(0)}v^{(4)}=u^{(4)}v+4u^{\prime\prime\prime}v’+6u^{\prime\prime}v^{\prime\prime}+4u’v^{\prime\prime\prime}+uv^{(4)}.$$
2. Llamemos $u=\sqrt{x}$ y $v=\log (x+1).$ Entonces,$$\begin{aligned}& u=x^{1/2},\;u(1)=1\\
   &u’=\frac{1}{2}x^{-1/2},\;u'(1)=\frac{1}{2}\\
   &u^{\prime\prime}=-\frac{1}{4}x^{-3/2},\;u»(1)=-\frac{1}{4}\\
   &u^{(3)}=\frac{3}{8}x^{-5/2},\;u^{(3)}(1)=\frac{3}{8}\\
   &u^{(4)}=-\frac{15}{16}x^{-7/2},\;u^{(4)}(1)=-\frac{15}{16}\\
   \end{aligned}\qquad
   \begin{aligned}& v=\log (x+1),\;v(1)=\log 2\\
   &v’=(x+1)^{-1},\;v'(1)=\frac{1}{2}\\
   &v^{\prime\prime}=-(x+1)^{-2},\;v»(1)=-\frac{1}{4}\\
   &v^{(3)}=2(x+1)^{-3},\;v^{(3)}(1)=\frac{1}{4}\\
   &v^{(4)}=-6(x+1)^{-4},\;v^{(4)}(1)=-\frac{3}{8}.
   \end{aligned}$$ Usando la fórmula de Leibniz: $$f^{(4)}(1)=\displaystyle\sum_{k=0}^4\binom{4}{k}u^{(4-k)}(1)v^{(k)}(1)$$ $$=-\dfrac{15}{16}\log 2+4\dfrac{3}{8}\dfrac{1}{2}+6\dfrac{-1}{4}\dfrac{-1}{4}+4\dfrac{1}{2}\dfrac{1}{4}-\dfrac{3}{8}$$ $$=-\dfrac{15}{16}\log 2+\dfrac{12}{16}+\dfrac{6}{16}+\dfrac{8}{16}-\dfrac{6}{16}$$ $$=\dfrac{20-15\log 2}{16}=\dfrac{5}{4}-\dfrac{15}{16}\log 2.$$
3. Llamemos $u=e^x$ y $v=\operatorname{sen}x$. Entonces, $$\begin{aligned}& u=e^x,\;u(\pi/2)=e^{\pi/2}\\
   &u’=e^x,\;u'(\pi/2)=e^{\pi/2}\\
   &u^{\prime\prime}=e^x,\;u^{\prime\prime}(\pi/2)=e^{\pi/2}\\
   &u^{(3)}=e^x,\;u^{(3)}(\pi/2)=e^{\pi/2}\\
   &u^{(4)}=e^x,\;u^{(4)}(\pi/2)=e^{\pi/2}\\
   \end{aligned}\qquad
   \begin{aligned}& v=\operatorname{sen}x,\;v(\pi/2)=1\\
   &v’=\cos x,\;v'(\pi/2)=0\\
   &v»=-\operatorname{sen}x,\;v»(\pi/2)=-1\\
   &v^{(3)}=-\cos x,\;v^{(3)}(\pi/2)=0\\
   &v^{(4)}=\operatorname{sen}x,\;v^{(4)}(\pi/2)=1.
   \end{aligned}$$ Usando la fórmula de Leibniz: $$f^{(4)}(\pi/2)=\displaystyle\sum_{k=0}^4\binom{4}{k}u^{(4-k)}(\pi/2)v^{(k)}(\pi/2)$$ $$=e^{\pi/2}+0+6e^{\pi/2}(-1)+0+e^{\pi/2}=-4e^{\pi/2}.$$
4. Llamando $u(x)=e^{\alpha x}$ y $v(x)=x^2$ obtenemos $$\left \{ \begin{matrix} u^{(0)}(x)=e^{\alpha x } \\ u^{(1)}(x)=\alpha e^{\alpha x }\\ u^{(2)}(x)=\alpha ^2 e^{\alpha x }\\u^{(3)}(x)=\alpha ^3 e^{\alpha x }\\ \ldots \\ u^{(n)}(x)=\alpha ^ne^{\alpha x }\end{matrix}\right.\;\;\; \left \{ \begin{matrix} v^{(0)}(x)=x^2 \\ v^{(1)}(x)=2x \\ v^{(2)}(x)=2\\ v^{(3)}(x)=0 \\ \ldots \\ v^{(n)}(x)=0\end{matrix}\right. $$ Entonces, $$(uv)^{(n)}(x)=\displaystyle\binom{n}{0}\alpha ^ne^{\alpha x }x^2+\displaystyle\binom{n}{1}\alpha ^{n-1}e^{\alpha x }\cdot{2x}+\displaystyle\binom{n}{2}\alpha ^{n-2}e^{\alpha x }\cdot{2}$$$$
   =\alpha ^ne^{\alpha x }x^2+2n\alpha ^{n-1}e^{\alpha x}x+n(n-2) \alpha ^{n-2}e^{\alpha x }\\
   = \alpha^{n-2}e^{\alpha x}(\alpha^2x^2+2n\alpha x+n^2-2n).$$
5. Tenemos $(uv)’=u’v+uv’.$ Derivando: $$\begin{aligned}&(uv)^{\prime\prime}=\left((uv)’\right)’=\left(u’v+uv’\right)’=u^{\prime\prime}v+u’v’+u’v’+uv^{\prime\prime}\\
   &=u^{\prime\prime}v+2u’v’+uv^{\prime\prime}=\displaystyle\binom{2}{0}u^{(2)}v^{(0)}+\displaystyle\binom{2}{1}u^{(1)}v^{(1)}+\displaystyle\binom{2}{2}u^{(0)}v^{(2)}.\end{aligned}$$ Derivando de nuevo: $$\begin{aligned}&(uv)^{\prime\prime\prime}=\left(u^{\prime\prime}v+2u’v’+uv^{\prime\prime}\right)’=u^{\prime\prime\prime}v+u^{\prime\prime}v’+2u^{\prime\prime}v’+2u’v^{\prime\prime}+u’v^{\prime\prime}+uv^{\prime\prime\prime}\\
   &=u^{\prime\prime\prime}v+3u^{\prime\prime}v’+3u’v^{\prime\prime}+uv^{\prime\prime\prime}\\
   &=\displaystyle\binom{3}{0}u^{(3)}v^{(0)}+\displaystyle\binom{3}{1}u^{(2)}v^{(1)}+\displaystyle\binom{3}{2}u^{(1)}v^{(2)}+\displaystyle\binom{3}{3}u^{(0)}v^{(3)}.\end{aligned}$$
6. La fórmula es cierta para $n=1,$ en efecto $$(uv)^{(1)}=(uv)’=u’v+uv’$$ $$=\displaystyle\binom{1}{0}u^{(1)}v^{(0)}+\displaystyle\binom{1}{1}u^{(0)}v^{(1)}=\sum_{k=0}^1\binom{1}{k}u^{(1-k)}v^{(k)}.$$
   Supongamos que la fórmula es cierta para $n.$ Entonces, $$(uv)^{(n+1)}=\biggl(\sum_{k=0}^n\binom{n}{k} u^{(n-k)} v^{(k)}\biggr)^{\prime}
   =\sum_{k=0}^n \left(\binom{n}{k} u^{(n-k)} v^{(k)}\right)^{\prime}$$ $$=\sum_{k=0}^n \binom{n}{k}\left(u^{(n-k+1)} v^{(k)}+u^{(n-k)} v^{(k+1)}\right)
   $$ $$=\sum_{k=0}^n \binom{n}{k}u^{(n-k+1)} v^{(k)}+\sum_{k=0}^n\binom{n}{k} u^{(n-k)} v^{(k+1)}$$ Haciendo un cambio de índices y usando las conocidas fórmulas combinatorias $$\binom{n}{k}+\binom{n}{k-1}=\binom{n+1}k,$$ $$\binom n0=1=\binom{n+1}0,\;\binom nn=1=\binom{n+1}{n+1}$$ podemos escribir $$(uv)^{(n+1)}=\sum_{k=0}^n \binom{n}{k}u^{(n-k+1)} v^{(k)}+\sum_{k=1}^{n+1}\binom{n}{k-1} u^{(n-k+1)} v^{(k)}$$ $$=\binom{n}{0}u^{(n+1)}v^{(0)}+\sum_{k=1}^n \binom{n}{k}u^{(n-k+1)} v^{(k)}$$ $$+\sum_{k=1}^{n}\binom{n}{k-1} u^{(n-k+1)} v^{(k)}+\binom{n}{n}u^{(0)}v^{(n+1)}$$ $$=\binom{n+1}{0}u^{(n+1)}v^{(0)}+\sum_{k=1}^n \binom{n+1}{k}u^{(n+1-k)} v^{(k)}+\binom{n+1}{n+1}u^{(0)}v^{(n+1)}$$ $$=\sum_{k=0}^{n+1} \binom{n+1}{k}u^{(n+1-k)} v^{(k)},$$ lo cual implica que la fórmula es cierta para $n+1.$
7. Llamando $u=e^x$ y $v=x$ obtenemos $$\left \{ \begin{matrix} u^{(0)}=e^{ x } \\ u^{(1)}= e^{ x }\\ u^{(2)}= e^{ x }\\u^{(3)}= e^{ x }\\ \ldots \\ u^{(n)}=e^{ x }\end{matrix}\right.\qquad \left \{ \begin{matrix} v^{(0)}=x \\ v^{(1)}=1 \\ v^{(2)}=0\\ v^{(3)}=0 \\ \ldots \\ v^{(n)}=0.\end{matrix}\right. $$ Por tanto, $$y^{(n)}=(uv)^{(n)}=\displaystyle\binom{n}{0}e^{ x }x+\displaystyle\binom{n}{1} e^{ x }\cdot{1}=xe^x+ne^x=(x+n)e^x.$$
8. Ver Derivada $f^{(2016)}(0)$ para $f(x)=x^2\log (x+1)$.