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Mathematics 19 Online
OpenStudy (studygurl14):

@solomonzelman @satellite73

OpenStudy (studygurl14):

Find the derivative @SolomonZelman @jdoe0001 @satellite73

OpenStudy (anonymous):

question is?

OpenStudy (studygurl14):

I know the formula is \(\huge \frac{1}{u\ln a}{\frac{du}{dx}}\), and u = \(\sqrt{x+1}\) But I don't know where to go from there.

OpenStudy (studygurl14):

a = 10

OpenStudy (anonymous):

oooh i see the question is "find the derivative"

OpenStudy (anonymous):

first off don't forget that \[\log_{10}(\sqrt{x+1})=\frac{1}{2}\log_{10}(x+1)\] that will make your life tons easier

OpenStudy (studygurl14):

OH!!!! I forgot about that rule...I think I can solve it now.

OpenStudy (anonymous):

so just make \(u=x+1\) and do it in your head

OpenStudy (anonymous):

if you have a bunch of time to waste you can do it your way and see that, after a bunch of algebra, you get the same thing but then again you could spend that time eating a snack

OpenStudy (studygurl14):

\(\huge \frac{1}{(2x+1)\ln 10}\)

OpenStudy (studygurl14):

is that right?

OpenStudy (anonymous):

hmm no

OpenStudy (anonymous):

the 2 should be a factor of the denominator

OpenStudy (studygurl14):

so then 2(x+1) instead of 2x + 2

OpenStudy (anonymous):

either way, but you wrote \(2x+1\)

OpenStudy (studygurl14):

Oh, oops. Well, I meant 2x + 2

OpenStudy (studygurl14):

So that's fine too?

OpenStudy (anonymous):

then that is fine with parentheses yes

OpenStudy (studygurl14):

awesome, thank you

OpenStudy (anonymous):

yw

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