If $f(x)=x\sqrt{2x3}$ what is $f'(x)$?
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Solution 1
Use product rule: $(uv)' = u'v + uv' $. In your problem
$u = x \implies u' = 1 $
$v = \sqrt{2x 3} \implies v' = \frac{2}{\sqrt{2x3}}$
Solution 2
You answer is close. There shouldn't be a square root in the second term since it's already to the 1/2 power. To simplify:
$$ \sqrt{2x3} + x(2x3)^{1/2} = \sqrt{2x3} + \frac{x}{\sqrt{2x3}} = \frac{(2x3)+x}{\sqrt{2x3}} = \frac{3x3}{\sqrt{2x3}} $$
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Author by
Hannah
Updated on August 01, 2022Comments

Hannah 3 months
so far I rewrote the problem using the product rule and chain rule so that i have $$\sqrt(2x3)+x(\sqrt(2x3)^{1/2}$$
Now what?

Eric Stucky over 8 yearsYou lost an $x$; that changes the methods you should use a bit.


Hannah over 8 yearsI ended up getting a common denominator to add them and I ended up with an answer that didn't make sense.

2012ssohn over 8 yearsYou should get $$(uv)' = u'v + uv' = 1\cdot\sqrt{2x3} + x\cdot\frac{2}{\sqrt{2x3}} = \frac{2x3}{\sqrt{2x3}} + \frac{2x}{\sqrt{2x3}} = \frac{4x3}{\sqrt{2x3}}$$

Hannah over 8 yearsThat is not an answer on my sheet so i'm not sure if there is a typo. And they did leave the radical in the denominator in all the answers so I don't think it's a simplification problem.

Dylan over 8 years$v' = \frac{1}{2}(2x3)^{1/2}⋅(2x)' = (2x3)^{1/2}$. There shouldn't be a $2$ in there.