Why must the power series for the derivative of this be shifted over one unit? I was asked to find the power series for $f(x) = \frac{x}{(1+7x)^2}$ I proceeded by finding the power series for  $g(x) = \frac{1}{1+7x}$, differentiating that series to get $g'(x) = \frac{-7}{(1+7x)^2}$, and mutiplying by $\frac{-x}{7}$ to get an equivalent answer. However, when looking at a solution to the problem, I saw that the power series for the power series for $g'(x)$ was shifted over by one unit to give:

$$\sum_{n=1}^\infty (-4)^nx^{n-1}$$

Why is the power series for $g'(x)$ shifted over here?
I don't see a problem.

$f(x)=\dfrac x{(1+7x)^2}=x(1+7x)^{-2}=x(1-14x+147x^2-...)=x-14x^2+147x^3-...$

$g(x)=\dfrac 1{1+7x}=1-7x+49x^2-343x^3...$

$g'(x)=-\dfrac 7{(1+7x)^2}=-7(1+7x)^{-2}=-7(1-14x+147x^2...)=-7+98x-1029x^2...$

$-\dfrac x7 g'(x)=f(x)=x-14x^2+147x^3....$
I don't see a problem.

$f(x)=\dfrac x{(1+7x)^2}=x(1+7x)^{-2}=x(1-14x+147x^2-...)=x-14x^2+147x^3-...$

$g(x)=\dfrac 1{1+7x}=1-7x+49x^2-343x^3...$

$g'(x)=-\dfrac 7{(1+7x)^2}=-7(1+7x)^{-2}=-7(1-14x+147x^2...)=-7+98x-1029x^2...$

$-\dfrac x7 g'(x)=f(x)=x-14x^2+147x^3....$