<s>[INST] 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
 [/INST] Induction is a very good way. For the induction step,
Say $P(n)$ is true for $n=k$ 
For $P(k+1)$, $$\frac{d}{dx^{k+1}}(9\sqrt x)$$
$$=\frac{d}{dx}\left(\frac{d}{dx^{k}}(9\sqrt x)\right)$$
$$=\frac{d}{dx} \left[(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (4x)^{\frac{1-2k}{2}}\right]$$
$$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{d}{dx} \left[(4x)^{\frac{1-2k}{2}}\right]$$
$$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{1-2k}{2}  (4x)^{\frac{-1-2k}{2}}\cdot 4$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1)  (4x)^{\frac{-1-2k}{2}}\cdot 2$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) \cdot \frac{2k}{k} \cdot (4x)^{\frac{-1-2k}{2}}$$
$$=(-1)^{[(k+1)-1]} \cdot \frac{9[2(k+1)-2]!}{[(k+1)-1]!} \cdot (4x)^{\frac{1-2(k+1)}{2}}$$
$P(k+1)$ is true.
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