Exercise:
For each natural number $n$ define $$\phi_n(x)=\frac{d^n}{d x^n}\left(x^2-1\right)^n$$ Derive the formulas $$\phi_{n+1}^{\prime}(x)=2(n+1) x \phi_n^{\prime}(x)+2(n+1)^2 \phi_n(x)$$ $$\phi_{n+1}^{\prime}(x)=\left(\left(x^2-1\right) \phi_n^{\prime}(x)\right)^{\prime}+2(n+1) x \phi_n^{\prime}(x)+(n+1)(n+2) \phi_n(x)$$ Hint. Attack the expressions $$\frac{d^{n+2}}{d x^{n+2}}\left(x^2-1\right)^{n+1} \space\space \frac{d^{n+1}}{d x^{n+1}}\left(x^2-1\right)^{n+1}$$
Mt attempt: $$\frac{d^{n+2}}{d x^{n+2}}\left(x^2-1\right)^{n+1}=\sum_{k=0}^{n+2}\binom{n+2}{k} \frac{d^{k}}{d x^{k}}\left(x^2-1\right) \frac{d^{n+2-k}}{d x^{n+2-k}}\left(x^2-1\right)^{n}$$ Obviously survives only $k=0,1,2$ - th terms $$\frac{d^{n+2}}{d x^{n+2}}\left(x^2-1\right)^{n+1}=(x^2-1)\frac{d^{n+2}}{d x^{n+2}}\left(x^2-1\right)^{n}+2x(n+2)\frac{d^{n+1}}{d x^{n+1}}\left(x^2-1\right)^{n}+(n+2)(n+1)\frac{d^{n+1}}{d x^{n+1}}\left(x^2-1\right)^{n}$$ $$\phi_{n+1}^{\prime}(x)=(x^2-1)\phi_n^{\prime\prime}(x)+2(n+2) x \phi_n^{\prime}(x)+(n+1)(n+2) \phi_n(x)$$ $$\phi_{n+1}^{\prime}(x)=\left(\left(x^2-1\right) \phi_n^{\prime}(x)\right)^{\prime}+2(n+1) x \phi_n^{\prime}(x)+(n+1)(n+2) \phi_n(x)$$ I succeeded to get the second formula but coludn't find the first formula.
Could anybody help me, please?
To simplify notation, I write $\delta_n$ instead of $\frac{d^n}{dx^n}$ ; thus $\phi_n(x)=\delta_n((x^2-1)^n)$. We then have
$$ \begin{array}{lcl} \phi_{n+1}(x) &=& \delta_{n+1} ((x^2-1)^{n+1}) \\ &=& \delta_{n+1} ((x^2-1)(x^2-1)^{n}) \\ &=& \delta_{n} (2x(x^2-1)^{n}+(x^2-1)2nx(x^2-1)^{n-1}) \\ &=& \delta_{n} (2x(x^2-1)^{n}+2nx(x^2-1)^{n}) \\ &=& 2(n+1)\delta_{n} (x(x^2-1)^{n}) \\ \end{array} $$
It follows that
$$ \begin{array}{lcl} \phi'_{n+1}(x) &=& 2(n+1)\delta_{n+1} (x(x^2-1)^{n}) \\ &=& 2(n+1)\sum_{j=0}^{n+1} \binom{n+1}{j} \delta_{j} (x) \delta_{(n+1)-j} ((x^2-1)^{n}) \\ &=& 2(n+1)\sum_{j=0}^{1} \binom{n+1}{j} \delta_{j} (x) \delta_{(n+1)-j} ((x^2-1)^{n}) \\ &=& 2(n+1)\bigg(x\delta_{n+1} ((x^2-1)^{n})+(n+1)\delta_{n} ((x^2-1)^{n})\bigg) \\ &=& 2(n+1)\bigg(x\phi'_{n}(x)+(n+1)\phi_n(x)\bigg) \\ \end{array} $$ which proves the first formula.