Proving formal power series $A(x) = \sum_{n = 0}^{\infty} a_nx^n$ with $A'(x) := \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$

Question

To the formal power series $A(x) = \sum_{n = 0}^{\infty} a_nx^n$ with the multiplicative inverse $A^{-1}(x)$ we define $A'(x) := \sum_{n=0}^{\infty}(n+1)a_{n+1}x^n$

How can one prove

$$(A^{-1})'(x) = - \frac{A'(x)}{A^2(x)}$$

I know that the statement is equivalent to $(A^{-1})(x)A(x) + A'(x)A^{-1}(x) = 0$

Since $A^{-1}(x)$ is the multiplicative inverse of $A$ it follows that for all $x$: $A(x) A^{-1}(x) = 1$

If we derive both sides to x (with using the product rule on the left side), then the equivalent statement is proven I think?

But how do I continue?

Answer 1

Differentiating $$A(x)A^{-1}(x)=1$$ we obtain \begin{align*} A^{\prime}(x)A^{-1}(x)+A(x)\left(A^{-1}(x)\right)^{\prime}=0\tag{1} \end{align*}

We obtain from (1) \begin{align*} \color{blue}{\left(A^{-1}(x)\right)^{\prime}}&=-\frac{A^{\prime}(x)A^{-1}(x)}{A(x)}\\ &\,\,\color{blue}{=-\frac{A^{\prime}(x)}{A^2(x)}} \end{align*}

The last step follows since $A^{-1}(x)=\frac{1}{A(x)}$.