$f^n(x) = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$ $f^n$(x)$ = \frac{5-10n}{2}*(7-5x)^{-1} * f^{n-1}(x)$. Prove valid for all $n \in \mathbb{N}$
Where $f^n(x)$ is the $n^{th}$ derivative of $f(x)$. And $f(x) = (7-5x)^{\frac{1}{2}}$.
So far I have made my base case of $n=1$ and showed that it equals the derivative of $f(x)$ and then I assumed n=k. I then state that: $f^{k+1} = \frac{d}{dx}(f^k)$. I know that my end goal is to get: $f^{k+1} = \frac{5-10(k+1)}{2}*(7-5x)^{-1} * f^{(k+1)-1}(x)$ which equals $f^{k+1} = \frac{-5-10k}{2}*(7-5x)^{-1} * f^{k}(x)$.
I've tried plugging the equation for $f^{k+1}$ into $f^{k+1} = \frac{d}{dx}(f^k)$ but couldn't seem to get the result I wanted, as I ended up with:
$-25*(7-5x)^{-2}*f^k(x)$
Help would be much appreciated.

The constant term in $f^n(x)$ should be $\frac{10n-15}{2}$, not $\frac{5-10n}{2}$.
It's best to rewrite the equation as
$$f^n(x) = a_n \frac{f^{n-1}(x)}{(f(x))^2}$$
This will greatly simplify our calculation.
Induction step from $n$ to $n+1$: First notice that
$$f^{n-1} (x) = \frac{1}{a_n} (f(x))^2 f^n(x)\tag1$$
and
$$f'(x) = a_1 \frac{f(x)}{(f(x))^2} \implies f(x)f'(x)= a_1\tag2$$
with $a_1=-\frac{5}{2}$.
Then
$$f^{n+1}(x) = \frac{d}{dx} f^n(x) = \frac{d}{dx} \left(a_n \cdot \frac{f^{n-1}(x)}{(f(x))^2}\right)\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - f^{n-1}(x) \cdot 2f(x)f'(x)}{(f(x))^4}\\
=a_n \cdot \frac{f^n(x) (f(x))^2 - \frac{1}{a_n} (f(x))^2 f^n(x) \cdot (2a_1)}{(f(x))^4}\\
= a_n \left(1-\frac{2a_1}{a_n}\right) \frac{f^n(x)}{(f(x))^2}\\
= (a_n-2a_1) \frac{f^n(x)}{(f(x))^2}
$$
Then we have $a_{n+1}=a_n-2a_1$, an arithmetic sequence. Therefore $a_n=(n-1)(-2a_1)+a_1 = \frac{10n-15}{2}.\blacksquare$
