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determine G''(4)
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I'm looking for the fastest way to solve this.
\[G(4)=0\]
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
0_o nevermind the question was determine G'(4)
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thanks though.
oh then that is much much easier!
think of it as \[G(x)=xF(x)\] so the derivative is \[G'(x)=F(x)+xF'(x)\] and \[G'(4)=F(4)+4F'(4)\]
where \(F\) is the integral, so \(F(4)=0\)
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yeah like that you are way ahead of me how did you get \(\arctan(\sqrt2 -1)\)
where?
it is \(\frac{\pi}{8}\) just wondering
oh nvm it is \(\arctan(1)\) doh ignore me
it is \[\arctan(\sqrt{4}-1)\]
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haha ok
yeah you are right i am senile today
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