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Mathematics 23 Online
OpenStudy (studygurl14):

calculus @solomonzelman

OpenStudy (studygurl14):

derivative of...

OpenStudy (studygurl14):

Should I pull out the exponent 2 first?

OpenStudy (xapproachesinfinity):

product and chain rule

OpenStudy (xapproachesinfinity):

you can if you did (sqrt6 x)^2

OpenStudy (solomonzelman):

\(\color{#000000 }{ \displaystyle 4x^5\ln(6x^2) }\) \(\color{#000000 }{ \displaystyle 8x^5\ln(6x) }\) \(\color{#000000 }{ \displaystyle 8x^5\ln(x)+8\ln(6)x^5 }\)

OpenStudy (xapproachesinfinity):

i would just use chain rule directly if it was me

OpenStudy (solomonzelman):

So, the second part is just power-rule, and first is just product... There are several ways to do the problem, tho'.

OpenStudy (xapproachesinfinity):

or like solo just did use log property of product

OpenStudy (studygurl14):

Oh, right...anyway, would the derivative of \(\large \ln(6x^2) = 2\Large \frac{1}{x^2}\)

OpenStudy (studygurl14):

sorry, ignore the set up there, I was solving in my head as I was typing the LaTeX

OpenStudy (xapproachesinfinity):

no x^2 at bottom is not good

OpenStudy (solomonzelman):

\(\color{#000000 }{ \displaystyle \ln(6x^2)=\frac{1}{6x^2}\times \left[\frac{d}{dx}~6x^2\right] }\)

OpenStudy (solomonzelman):

using the chain rule.

OpenStudy (studygurl14):

Ah, so 2/x?

OpenStudy (solomonzelman):

precisely;)

OpenStudy (xapproachesinfinity):

yes

OpenStudy (studygurl14):

Thank you both!

OpenStudy (xapproachesinfinity):

welcome

OpenStudy (solomonzelman):

Just to verify, if you want, you can post the final answer for check.

OpenStudy (solomonzelman):

or, you can check it through wolf..

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