Prove n-th derivative function Let $f(x)=(3x+5)^{1/2}$. Obtain and prove a formula for the $n$-th derivative $f^{(n)}$
I may need to find some derivatives of the function, and prove by induction, but I do not know how to do these.
F’(x) = 3/2(3x+5)^(-1/2)
F''(x) = -9/4(3x+5)^(-3/2)
F'''(x) = 81/8(3x+5)^(-5/2)
F''''(x) = -1215/16(3x+5)^(-7/2)
Pattern:
F^(n)(x)=(1/2−(n−1))(1/2−(n−2))...(1/2−(n−n))3^n(3x+5)(1/2−n)
These repeated steps ...(1/2−(n−n)) continue until the number(n - number) reaches n value. E.g. if n is 1 then F'(x)=(1/2-(1-1))(3x+5)^(-1/2)

Let
$f(x)
=(ax+b)^c
$.
What is
$f^{(n)}(x)
$?
Each derivative
decreases the exponent
by 1.
So,
let
$f^{(n)}(x)
=g(n)(ax+b)^{c-n}
$.
Then
$f^{(n+1)}(x)
=g(n)(c-n)a(ax+b)^{c-n-1}
$,
so that
$g(n+1)
=g(n)(c-n)a
$,
or
$g(n+1)/g(n)
=(c-n)a
$.
Since
$g(0) = 1$,
$\begin{array}\\
g(n)
&=\prod_{k=0}^{n-1} g(k+1)/g(k)\\
&=\prod_{k=0}^{n-1} (c-k)a\\
&=a^{n}\prod_{k=0}^{n-1} (c-k)\\
&=a^{n}\Gamma(c+1)/\Gamma(c-n+1)\\
&=a^{n}(c)!/(c-n)!
\qquad\text{if } c \text{ is an integer}\\
\end{array}
$
If
$c = u/v$,
$\begin{array}\\
g(n)
&=a^{n}\prod_{k=0}^{n-1} (c-k)\\
&=a^{n}\prod_{k=0}^{n-1} (u/v-k)\\
&=a^{n}\prod_{k=0}^{n-1} (u-kv)/v\\
&=(a/v)^{n}\prod_{k=0}^{n-1} (u-kv)\\
\end{array}
$
If $c=\frac12$,
so $u=1, v=2$,
$\begin{array}\\
g(n)
&=\frac{a^n}{2^n}\prod_{k=0}^{n-1} (1-2k)\\
&=\frac{(-1)^{n}a^n}{2^n}\prod_{k=0}^{n-1} (2k-1)\\
&=\frac{(-1)^{n+1}a^n}{2^n}\prod_{k=1}^{n-1} (2k-1)\\
&=\frac{(-1)^{n+1}a^n}{2^n}\prod_{k=1}^{n-1} (2k-1)\frac{2k}{2k}\\
&=\frac{(-1)^{n+1}a^n}{2^n}\frac{\prod_{k=1}^{n-1} (2k-1)(2k)}{\prod_{k=1}^{n-1}(2k)}\\
&=\frac{(-1)^{n+1}a^n}{2^n}\frac{(2n-2)!}{2^{n-1}(n-1)!}\\
&=\frac{(-1)^{n+1}a^n}{2^{2n-1}}\frac{(2n-2)!}{(n-1)!}\\
\end{array}
$
