Prove n-th derivative function

Question

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)

Answer 1

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} $