Let y = f(x) possess derivatives of all orders. Prove

Let $y = f(x)$ possess derivatives of all orders. Prove that \[ y'\sum_{v=1}^{n}\bigl(-1\bigr)^{v}\binom{n}{v}y^{-v}\frac{d^{n}\bigl(y^{v}\bigr)}{dx^{n}} = \frac{y}{n+1}\sum_{v=1}^{n+1}\bigl(-1\bigr)^{v+1}\binom{n+1}{v}y^{-v}\frac{d^{n+1}\bigl(y^{v}\bigr)}{dx^{n+1}}. \tag{$*$} \].

Step 1: Strategy

Rather than transforming one side into the other, evaluate the alternating sum once and for all. Define

\[ G_{n} := \sum_{v=0}^{n}\bigl(-1\bigr)^{v}\binom{n}{v}y^{-v}\bigl(y^{v}\bigr)^{(n)}. \]

For $n\ge1$ the term $v = 0$ contributes nothing, since $y^{0} = 1$ has zero derivatives; so $G_{n}$ is exactly the sum appearing in $(*)$.

Step 2: The key lemma

\[ G_{n} = \bigl(-1\bigr)^{n}\,n!\left(\frac{y'}{y}\right)^{n}. \tag{1} \]

Two immediate checks: $G_{1} = -\dfrac{y'}{y}$, and

\[ G_{2} = -\frac{2y''}{y}+\frac{\bigl(y^{2}\bigr)''}{y^{2}} = -\frac{2y''}{y}+\frac{2\bigl(y'\bigr)^{2}+2yy''}{y^{2}} = 2\left(\frac{y'}{y}\right)^{2}, \]

both matching (1).

Step 3: Proof of the lemma

Write $y = e^{g}$ with $g = \ln y$, so that $y^{v} = e^{vg}$. By Faà di Bruno's formula, $\bigl(e^{vg}\bigr)^{(n)} = e^{vg}\sum_{k=1}^{n}B_{n,k}\bigl(g',g'',\ldots\bigr)v^{k}$.

where $B_{n,k}$ are the partial Bell polynomials. Dividing by $y^{v} = e^{vg}$ and substituting: \[G_{n} = \sum_{k=1}^{n}B_{n,k}\sum_{v=0}^{n}\bigl(-1\bigr)^{v}\binom{n}{v}v^{k}. \tag{2}\]

Step 4: The inner sum kills all but one term

The inner sum is an $n$-th finite difference of $v^{k}$, and

\[ \sum_{v=0}^{n}\bigl(-1\bigr)^{v}\binom{n}{v}v^{k} = \bigl(-1\bigr)^{n}\,n!\,S(k,n), \]

with $S(k,n)$ the Stirling numbers of the second kind. Since $S(k,n) = 0$ for $k\lt n$ and $S(n,n) = 1$, only $k = n$ survives in (2): \[G_{n} = \bigl(-1\bigr)^{n}\,n!\,B_{n,n}\bigl(g',g'',\ldots\bigr) = \bigl(-1\bigr)^{n}\,n!\bigl(g'\bigr)^{n},\]

because the top Bell polynomial is $B_{n,n} = \bigl(g'\bigr)^{n}$. With $g' = \dfrac{y'}{y}$, this is (1).

Step 5: Evaluate the left side of $(*)$

\[ \text{LHS} = y'\,G_{n} = \bigl(-1\bigr)^{n}\,n!\,\frac{\bigl(y'\bigr)^{n+1}}{y^{n}}. \tag{3} \]

Step 6: Evaluate the right side of $(*)$

The sign $\bigl(-1\bigr)^{v+1}$ is the negative of $\bigl(-1\bigr)^{v}$, so the sum on the right is $-G_{n+1}$. Applying (1) at order $n+1$: \[-G_{n+1} = -\bigl(-1\bigr)^{n+1}\bigl(n+1\bigr)!\left(\frac{y'}{y}\right)^{n+1} = \bigl(-1\bigr)^{n}\bigl(n+1\bigr)!\,\frac{\bigl(y'\bigr)^{n+1}}{y^{n+1}},\]

\[ \text{RHS} = \frac{y}{n+1}\cdot\bigl(-1\bigr)^{n}\bigl(n+1\bigr)!\,\frac{\bigl(y'\bigr)^{n+1}}{y^{n+1}} = \bigl(-1\bigr)^{n}\,n!\,\frac{\bigl(y'\bigr)^{n+1}}{y^{n}}. \tag{4} \]

Step 7: Verification

\[n = 1:\ \text{LHS} = y'\left(-\frac{y'}{y}\right) = -\frac{\bigl(y'\bigr)^{2}}{y}; \quad \text{RHS} = \frac{y}{2}\left[\frac{2y''}{y}-\frac{\bigl(y^{2}\bigr)''}{y^{2}}\right] = -\frac{\bigl(y'\bigr)^{2}}{y} \;\;\checkmark\]

\[n = 2:\ \text{LHS} = 2\frac{\bigl(y'\bigr)^{3}}{y^{2}}; \quad \text{RHS} = \frac{y}{3}\left[\frac{3y'''}{y}-\frac{3\bigl(y^{2}\bigr)'''}{y^{2}}+\frac{\bigl(y^{3}\bigr)'''}{y^{3}}\right] = 2\frac{\bigl(y'\bigr)^{3}}{y^{2}} \;\;\checkmark\]

\[\text{Symbolic check for } n = 1,2,3:\ \text{lemma residual and LHS}-\text{RHS both vanish identically} \;\;\checkmark\]