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Mathematics 17 Online
OpenStudy (studygurl14):

How to find the derivative of @dan815 @hartnn @freckles

OpenStudy (studygurl14):

\(\large g(x)=\Large\sqrt{\frac{x+3}{2}}\)

hartnn (hartnn):

the denominator \(\sqrt 2\) is a constant, you can bring that out. just try to get the derivative of \(\sqrt{x+3}\)

OpenStudy (studygurl14):

Oh, okay. So would I change it to \(\Large (x+3)^{1/2}\)

OpenStudy (studygurl14):

And use the chain formula?

hartnn (hartnn):

yes, now you'll have to use the chain rule, and the fact that \(\Large (x^n)' = n x^{n-1}\)

OpenStudy (studygurl14):

So, would it be \(\Large\frac{1}{2\sqrt{x+3}}\)

hartnn (hartnn):

right and finally put the \(\sqrt 2\) in the denominator

OpenStudy (studygurl14):

Okay, so then \(\Large\frac{1}{2\sqrt{2x+6}}\)

hartnn (hartnn):

right or \(\Large\frac{1}{2\sqrt{2(x+3)}}\)

OpenStudy (studygurl14):

Is there a preferred way?

hartnn (hartnn):

whatever you feel is more simplified

OpenStudy (studygurl14):

Okay, thank you

hartnn (hartnn):

welcome ^_^

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