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Mathematics 9 Online
OpenStudy (anonymous):

determine G''(4)

OpenStudy (anonymous):

OpenStudy (anonymous):

I'm looking for the fastest way to solve this.

OpenStudy (anonymous):

\[G(4)=0\]

OpenStudy (anonymous):

but that doesn't mean \(G''(4)=0\) you have to actually that the derivative in this case using the product rule and the chain rule

OpenStudy (anonymous):

0_o nevermind the question was determine G'(4)

OpenStudy (anonymous):

thanks though.

OpenStudy (anonymous):

oh then that is much much easier!

OpenStudy (anonymous):

think of it as \[G(x)=xF(x)\] so the derivative is \[G'(x)=F(x)+xF'(x)\] and \[G'(4)=F(4)+4F'(4)\]

OpenStudy (anonymous):

where \(F\) is the integral, so \(F(4)=0\)

OpenStudy (anonymous):

OpenStudy (anonymous):

yeah like that you are way ahead of me how did you get \(\arctan(\sqrt2 -1)\)

OpenStudy (anonymous):

where?

OpenStudy (anonymous):

it is \(\frac{\pi}{8}\) just wondering

OpenStudy (anonymous):

oh nvm it is \(\arctan(1)\) doh ignore me

OpenStudy (anonymous):

it is \[\arctan(\sqrt{4}-1)\]

OpenStudy (anonymous):

haha ok

OpenStudy (anonymous):

yeah you are right i am senile today

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