# If $f(x)=x\sqrt{2x-3}$ what is $f'(x)$?

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## Solution 1

Use product rule: $(uv)' = u'v + uv'$. In your problem

$u = x \implies u' = 1$

$v = \sqrt{2x -3} \implies v' = \frac{2}{\sqrt{2x-3}}$

## 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:

$$\sqrt{2x-3} + x(2x-3)^{-1/2} = \sqrt{2x-3} + \frac{x}{\sqrt{2x-3}} = \frac{(2x-3)+x}{\sqrt{2x-3}} = \frac{3x-3}{\sqrt{2x-3}}$$

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### Hannah

Updated on August 01, 2022

• Hannah 3 months

so far I re-wrote the problem using the product rule and chain rule so that i have $$\sqrt(2x-3)+x(\sqrt(2x-3)^{-1/2}$$

Now what?

• Eric Stucky over 8 years
You lost an $x$; that changes the methods you should use a bit.
• Hannah over 8 years
I ended up getting a common denominator to add them and I ended up with an answer that didn't make sense.
• 2012ssohn over 8 years
You should get $$(uv)' = u'v + uv' = 1\cdot\sqrt{2x-3} + x\cdot\frac{2}{\sqrt{2x-3}} = \frac{2x-3}{\sqrt{2x-3}} + \frac{2x}{\sqrt{2x-3}} = \frac{4x-3}{\sqrt{2x-3}}$$
• Hannah over 8 years
That 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
$v' = \frac{1}{2}(2x-3)^{-1/2}⋅(2x)' = (2x-3)^{-1/2}$. There shouldn't be a $2$ in there.