<s>[INST] Product Chain Power Rule： Either it's the book or I am wrong. The problem:　$x^3\sqrt{2x+4}$ 
$f(x):= x^3$,  $g(x):= \sqrt{2x+4}$   
$(f\times g)' = f^{\prime}g+fg^{\prime}$  thus it should be 
$3x^2\sqrt{2x+4} + (x^3)[\frac{1}{2}(2x+4)^{\frac{-1}{2}}(2)]$ 
which is $3x^2\sqrt{2x+4}+\frac{x^3}{\sqrt{2x+4}}$ 
The book gives: $3x^2\sqrt{2x+4}+\frac{x^3}{2\sqrt{2x+4}}$ 
I'm correct? I always get worry when my answers don't match the book.  
$\frac{d}{dx}[f\times g(h(x))] = f^{\prime} \times g(h(x))+ f\times g^{\prime}(h(x))h^{\prime}$ right? 
 [/INST] As my colleagues have astutely pointed out, the product rule states $(fg)^{\prime} = f^{\prime} g + f g^{\prime}$. Define $f(x)= x^3$ and $g(x)= \sqrt{2x+4}$. As $f^{\prime}(x) = 3 x^{2}$ and $g^{\prime}(x) = \frac{1}{\sqrt{2x+4}}$,  the product rule gives
\begin{align}
(f(x)g(x))^{\prime} = f^{\prime}(x) g(x) + f(x) g^{\prime}(x) = 3x^{2} \sqrt{2x + 4} + \frac{x^3}{\sqrt{2x+4}}. 
\end{align}
The answer in your book has an incorrect factor of $2$ in the denominator of the second term. This factor should cancel with the factor of $2$ coming from $2x$ (in $g(x)$) by the chain rule.
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