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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?

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Last formula should be $g'(h(x))h'$, not $g(h(x))'h'$. – Arturo Magidin Feb 21 '12 at 2:55
Your answer to the problem is the correct one, I think. Books have typos. For simple derivatives such as this you can check with a computer, if you trust computers. – Dylan Moreland Feb 21 '12 at 2:55
@DylanMoreland I really know that I am correct, its just that I panic when I don't have what the book says. – yiyi Feb 21 '12 at 2:57
@Dylan You could write an answer so that OP will choose the answer and keep this away from those Unanswered list. – user21436 Feb 21 '12 at 3:33

2 Answers 2

up vote 2 down vote accepted

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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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 for instance, your answer can be verified by simplifying the expression.

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