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Mathematics 14 Online
OpenStudy (anonymous):

Differenciate y= (2sin(x+1))^1/2, show steps

OpenStudy (anonymous):

http://www.twiddla.com/589515 come here

OpenStudy (anonymous):

dy/dx=(1/2)(2cos(x+1).(1))^(-1/2) dy/dx=1/(cos(x+1))^(1/2)

myininaya (myininaya):

\[y=(f(g(x)) => y'=g'(x)f'(g(x))\] thats not right bell

OpenStudy (anonymous):

i just worked it out using the chain rule lol

myininaya (myininaya):

\[y'=\frac{1}{2}(2\sin(x+1))^{\frac{1}{2}-1}(2\sin(x+1))'\] \[y'=\frac{1}{2}(2\sin(x+1))^{\frac{-1}{2}}*2(\cos(x+1))\]

OpenStudy (anonymous):

myininaya: huhu.. thanks 4 correct me.. after this i won't do wrong again..

OpenStudy (anonymous):

yep, that is the correct answer LOL

myininaya (myininaya):

lol dont worry we all make mistakes

myininaya (myininaya):

i gave you a medal bellz :)

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