<s>[INST] $n^{th}$ derivative of $y=(x \sqrt{1+x^2})^m$ I was trying to find the $n^{th}$ derivative and $y_n(0)$(nth derivative at zero) of the function 
$$y=(x \sqrt{1+x^2})^m$$ 
Here is my approach:
$$y^2=(x^2 (1+x^2)^m$$
$$2yy_1=m(x^2 (1+x^2))^{m-1} \cdot(2x+4x^3)$$
$$2yy_1=m\frac{(x^2 (1+x^2))^{m}}{(x^2 (1+x^2)} \cdot(2x+4x^3)$$
$$2yy_1=m\frac{y^2}{(x^2 (1+x^2)} \cdot(2x+4x^3)$$
$$2yy_1((x^2 (1+x^2))=my^2(2x+4x^3)$$
$$y_1(x^2(1+x^2))=my(x+2x^3)$$
Derivating again I can write
$$y_2(x(1+x^2))+y_1(1+3x^2))=my_1(x+2x^3) + my(1+6x^2)$$
$$y_2(x(1+x^2))+y_1((1+3x^2)-m(x+2x^3))-my(1+6x^2)=0$$
Now my book says that the answer is 
$$y_n(0)=m^2(m^2-2^2)(m^2-4^2).......[m^2-(2n-2)^2]; n=odd$$
$$y_n(0)=m^2(m^2-1^2)(m^2-3^2).......[m^2-(2n-1)^2]; n=even$$
But after applying Leibnitz theorem for the derivative, I don't end up with anything like this. Can anybody please tell me how to approach this question.
 [/INST] You can use the series expansion:
$$
(1+x)^\alpha=\sum_{k=0}^\infty\pmatrix{\alpha \\ k}x^k
$$
where
$$
\pmatrix{\alpha \\ k}=\frac{\alpha\cdot(\alpha-1)\cdot\ldots\cdot(\alpha-k+2)\cdot(\alpha-k+1)}{k\cdot(k-1)\cdot\ldots\cdot2\cdot1}.
$$
Then, if $m$ is a positive integer,
$$
\left(x\sqrt{1+x^2}\right)^m=x^m\sum_{k=0}^\infty\pmatrix{m/2 \\ k}x^{2k}
=\sum_{k=0}^\infty\pmatrix{m/2 \\ k}x^{2k+m}
$$
Equating to the McLaurin series, we have, if $m$ is even
$$
y^{(n)}(0)=\begin{cases}
n!\pmatrix{m/2 \\ (n-m)/2} && n\ \text{even and}\ n\geq m\\
0 && n\ \text{odd}
\end{cases}
$$
and if $m$ is odd
$$
y^{(n)}(0)=\begin{cases}
0 && n\ \text{even} \\
n!\pmatrix{m/2 \\ (n-m)/2} && n\ \text{odd and}\ n\geq m
\end{cases}
$$
</s>