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?
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.
Your answer is correct and the best way to check that you are indeed correct and book is wrong is by using Wolfram Alpha, check http://www.wolframalpha.com/input/?i=derivative+of+x%5E3%5Csqrt%7B2x%2B4%7D for instance, your answer can be verified by simplifying the expression.
