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Differenciate y= (2sin(x+1))^1/2, show steps
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dy/dx=(1/2)(2cos(x+1).(1))^(-1/2) dy/dx=1/(cos(x+1))^(1/2)
\[y=(f(g(x)) => y'=g'(x)f'(g(x))\] thats not right bell
i just worked it out using the chain rule lol
\[y'=\frac{1}{2}(2\sin(x+1))^{\frac{1}{2}-1}(2\sin(x+1))'\] \[y'=\frac{1}{2}(2\sin(x+1))^{\frac{-1}{2}}*2(\cos(x+1))\]
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myininaya: huhu.. thanks 4 correct me.. after this i won't do wrong again..
yep, that is the correct answer LOL
lol dont worry we all make mistakes
i gave you a medal bellz :)
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