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Mathematics 21 Online
OpenStudy (anonymous):

Find the Derivative:

OpenStudy (anonymous):

You can use Equation Editor too!

OpenStudy (anonymous):

come on post the problem

OpenStudy (anonymous):

\[y=(x ^{2}+2)sort{x+2}\]

OpenStudy (anonymous):

and by sort it means "square root of x+2"

OpenStudy (anonymous):

\[y = (x^2 + 2) \times \sqrt{x+2}\]

OpenStudy (anonymous):

mhm ^ perfect...

OpenStudy (anonymous):

\[\frac{dy}{dx} = 2x \times \sqrt{x+2} + \frac{1}{2}\times x^2 \times \frac{1}{\sqrt{x+2}} + 1 \times \frac{1}{\sqrt{x+2}}\]

OpenStudy (liizzyliizz):

val!? haha

OpenStudy (anonymous):

looks like a product rule problem to me. use \[(fg)'=f'g+g'f\] with \[f(x)=(x^2+2),f'(x)=2x,g(x)=\sqrt{x+2},g'(x)=\frac{1}{2\sqrt{x+2}}\]

OpenStudy (anonymous):

or if you want a snap method, you could write \[(x+2)^2\times \sqrt{x+2}=(x+2)^{\frac{5}{2}}\] and use the power rule

OpenStudy (liizzyliizz):

val which problem are you doing o.o isnt it \[(x^2-4)\sqrt{x+2}\]

OpenStudy (liizzyliizz):

nvm lol

OpenStudy (anonymous):

mhm.. liz..

OpenStudy (anonymous):

you all working on the same homework?

OpenStudy (anonymous):

Thank you :)

OpenStudy (anonymous):

Yup

OpenStudy (liizzyliizz):

were in the same class lol and i asked this question earlier but the person helping me did not help o.o i was wondering why i was missing part of the answer when i checked it on the calculator

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