<s>[INST] Finding the second derivative of$f(x)=x^2\sqrt{4-x}$ 
Find the second derivative of the function following:

$$f(x)=x^2\sqrt{4-x}$$
Here I go...
$$f(x)= x^2(4-x)^{1\over 2}$$
\begin{align*}
f'(x) &= 2x(4-x)^{1\over 2}+{1\over 2}x^2(4-x)^{-{1\over 2}}(-1)\\ 
&= 2x(4-x)^{1\over 2} - {1 \over 2}x^2(4-x)^{-{1\over 2}}\\ 
&={1\over 2}x(4-x)^{-{1\over 2}}[4(4-x)-x]\\ 
&= {1\over 2}x(4-x)^{-{1\over 2}}(16-5x)\\
f''(x)&= {-{1\over 4}}x(4-x)^{-{3\over 2}}(-1)({1\over 2})(16-5x)+(-5)[{1\over 2}x(4-x)^{-{1\over 2}}]\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}(16-5x)-{5\over 2}(4-x)^{-{1\over 2}}\\
&= {-{1\over 8}}x(4-x)^{-{3\over 2}}[(16-5x)-20(4-x)]\\
&={-{1\over 8}}x(4-x)^{-{3\over 2}}(-64+15x^2)
\end{align*}
I think I messed up on the second derivative. Could anyone show me the steps for the second derivative? Thanks!
 [/INST] Notice that
$$f'(x)=\frac12(4-x)^{-1/2}(16x-5x^2)$$
Then
\begin{align*}
f''(x)&=\frac14(4-x)^{-3/2}(16x-5x^2)+(4-x)^{-1/2}(8-5x)\\
&=\frac{4(4-x)(8-5x)+(16x-5x^2)}{4(4-x)^{3/2}}\\
&=\frac{15x^2-96x+128}{4(4-x)^{3/2}}
\end{align*}
</s>