$n^{th}$ derivative of $\sqrt{ax+b}$?

Question

What is the $n^{th}$ derivative of $\sqrt{ax+b}$?

I don't understand where does $1.3.5.....(2n-1)$ factorial comes from. Except this

I have understood the whole question.

Please help me to solve this problem.I shall be very thankful for the help.

Answer 1

Hint: Write $$f(x)=(ax+b)^{1/2}$$ then $$f'(x)=\frac{1}{2}(ax+b)^{-1/2}a$$ $$f''(x)=\frac{a}{2}(ax+b)^{-3/2}a=…$$ Can you proceed?

Answer 2

The successive derivatives of a power go like

$$x^r\to rx^{r-1}\to r(r-1)x^{r-2}\to r(r-1)(r-2)x^{r-3}\to\cdots$$

Now if $r$ is a half-integer, semi-factorials appear.

Ignoring the signs,

$$\frac12\frac12\frac32\frac52\frac72\cdots$$

PS: the same holds for $(ax+b)^r$, provided you add a factor $a$ each time.