To find the nth Derivative of $9\sqrt{x}$ Can you help me to proof that the nth derivative of $9\sqrt{x}$ is $$ (-1)^{(n-1)} \cdot \frac{9(2n-2)!}{(n-1)!} \cdot (4x)^{\frac{1-2n}{2}}$$
I've tried induction but didn't go very far.
Many thanks

Firstly, the 9 is unimportant. Induction is the correct way to proceed, but perhaps we can disguise it in a computation: 
$$\begin{align} \frac{d^n}{dx^n} \left( x^{\frac{1}{2}} \right) &= \frac{d^{n-1}}{dx^{n-1}} \left( \frac{1}{2} x^{\frac{1}{2} -1} \right) \\&= \frac{d^{n-2}}{dx^{n-2}}\left( \frac{1}{2} \left(\frac{1}{2} - 1 \right) x^{\frac{1}{2} -2} \right) \\&= \dots =\frac{d^{n-k}}{dx^{n-k}} \left( \frac{1}{2} \left( \frac{1}{2} -1\right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} -(k-1) \right) x^{\frac{1}{2} - k}\right) \\&= \frac{1}{2} \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} - (n-1) \right) x^{\frac{1}{2} -n }. \end{align} $$ Can you simplify this?
