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Mathematics 19 Online
OpenStudy (lgbasallote):

LGBADERIVATIVE \[\huge y = \sqrt{1+2e^{3x}}\]

OpenStudy (lgbasallote):

\[y^2 = 1+2e^{3x}\]

OpenStudy (helder_edwin):

no need for that

OpenStudy (lgbasallote):

\[2yy\ = 1 + 6e^{3x}\]

OpenStudy (lgbasallote):

i mean \[2yy' = 1 + 6e^{3x}\]

OpenStudy (lgbasallote):

\[y' = \frac{1 + 6e^{3x}}{2\sqrt{1 + 2e^{3x}}}\]

OpenStudy (lgbasallote):

right?

OpenStudy (helder_edwin):

just apply tha chain rule: \[ \Large y'=\frac{1}{2\sqrt{1+2e^{3x}}}2e^{3x}3 \]

OpenStudy (zarkon):

no

OpenStudy (lgbasallote):

no? why?

OpenStudy (lgbasallote):

oh lol derivative of 1 is 0

OpenStudy (zarkon):

\[\frac{d}{dx}[1]=0\]

OpenStudy (lgbasallote):

*facepalm*

OpenStudy (turingtest):

hail Zarkon!

OpenStudy (lgbasallote):

i cant believe i make noobish mistakes :/ i really havent mastered derivatives lol

OpenStudy (zarkon):

yeah to me..I know the derivative of 1 is 0 :)

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