I have to find a formula for $f^{(n)}(x)$ for any $n \in N$ where $f(x) = \sqrt{4x-5}$. I've gotten this far:
${(-1)^{n-1} * something}\over(4x -5)^{{2n-1}\over2}$
After taking the first, second, and third derivatives, their numerators have been: 2, -4, 24, 240, 3360. A hint was given for this question, which was to double factorials (n!!). However, even with that I still find myself stuck.
A hint: calculate the first few derivatives without simplifying anything:
$$f^{(0)}(x)=(4x-5)^{1/2}$$
$$f^{(1)}(x)=\frac{1}{2}(4x-5)^{-1/2}\cdot4$$
$$f^{(2)}(x)=\frac{1}{2}\cdot\frac{-1}{2}(4x-5)^{-3/2}\cdot4^2$$
$$f^{(3)}(x)=\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}(4x-5)^{-5/2}\cdot4^3$$
$$f^{(4)}(x)=\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}(4x-5)^{-7/2}\cdot4^4$$
Can you extrapolate and put this into a pattern?
More hinting: $$\begin{align} f^{(5)}(x)&=\frac{1}{2}\cdot\frac{-1}{2}\cdot\frac{-3}{2}\cdot\frac{-5}{2}\cdot\frac{-7}{2}(4x-5)^{-9/2}\cdot4^5\\ &=(-1)^4\frac{1\cdot\color{blue}{2}\cdot3\cdot\color{blue}{4}\cdot5\cdot\color{blue}{6}\cdot7\cdot\color{blue}{8}}{2\cdot2^4\cdot\color{blue}{2}\cdot\color{blue}{4}\cdot\color{blue}{6}\cdot\color{blue}{8}}(4x-5)^{-9/2}\cdot4^5\\ &=(-1)^4\frac{1\cdot2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8}{2\cdot2^4\cdot2^4\cdot1\cdot2\cdot3\cdot4}(4x-5)^{-9/2}\cdot4^5\\ \end{align}$$