I've been struggling with this problem. I'm asked to find a recursive formula for the derivatives of this function: $$f(x) = \frac{1}{a+bx}$$ and then to prove the formula I've found with induction.
Attempted solution:
By differentiating this formula multiple times I've found that (we'll call this formula (1)): $$f^n(x) = (-1)^n(n!)b^n(a+bx)^{-(n+1)} $$
Or alternatively (we'll call this formula (2)): $$f^n(x)= (-nb)f^{n-1}(x)*f^0(x) $$
Which formula is the correct one? I think it's (2). I also have trouble proving those with induction. I can only prove one by using them both. That's what I've done:
if $n=1$, basically $$f'(x) = -b(a+bx)^{-1}$$ for $n+1$ we should find that: $$f^{n+1}(x)= (-1)^{n+1}(n+1)!b^{n+1}(a+bx)^{-n-2}$$ we'll call this formula (3) We also know that: $f^{n+1}(x) = \frac {d}{dx} f^n(x)$, so, using (2): $$f^{n+1}(x) = \frac {d}{dx} ((-nb)f^{n-1}(x)*f^0(x)) = (-nb)(f^{n}(x)*f^0(x)+f^{n-1}*f^1(x))$$ Then, using (1): $$f^{n+1}(x) =(-nb)*[(-1)^n(n!)b^n(a+bx)^{-n-1}*(a+bx)^{-1} + (-1)^{n-1}(n-1)!b^{n-1}(a+bx)^{-n-2}*(a+bx)^{-2}(-b) ] =$$
$$=(-nb)[(-1)^n(b^n)(a+bx)^{-n-2}(n-1)!(n+1)]=(-1)^{n+1}(n+1)!(b^{n+1})(a+bx)^{-n-2}$$
Which is exactly (3). Is my method correct? I think it's a bit messy and I've only succeeded in proving (1) when I wanted to prove (2) too.
Both of your formulas are correct, but if you want something recursive, you number 2 is what you're looking for: $$f^n(x)= (-nb)f^{n-1}(x)*f(x) \tag{2}$$ (Here I've used $f(x) \equiv f^0(x)$).
For your induction step, assuming that $f^n(x)= (-nb)f^{n-1}(x)*f(x)$, you have: \begin{align} f^{n + 1} &:= \frac{d}{dx}f^n(x)\\ &= \frac{d}{dx}[(-nb)f^{n-1}(x)*f(x)]\ \text{ (by the I.H)}\\ &= (-nb)(f^n(x)f(x) + f^{n - 1}f'(x)) \end{align} Use the induction hypothesis (I.H) again on $f'(x)$ to get $f'(x) = (-(1)b)f(x)f(x)$. Then, \begin{align} f^{n + 1}(x) &= (-nb)f^n(x)f(x) + (-nb)(-b)f^{n - 1}f(x)f(x) \end{align} Once again, use the I.H to substitute $(-nb)f^{n - 1}f(x) = f^n(x)$. Finally, \begin{align} f^{n + 1}(x) &= (-nb)f^n(x)f(x) + (-b)f^n(x)f(x)\\ &= (-(n + 1)b)f^nf(x) \end{align} as desired.