What is the derivative of this equation: y= arcsin (radical(2t))
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The chain rule is For your function make so Now find and
hi just to add,
Then what is the derivative of (2t^.5)
When i took the derivative of (2t^.5) i got 1/(2t^.5) does that even make sense?
Originally Posted by emilyroy10 Then what is the derivative of (2t^.5)
what is the derivative of ((2t)^.5)
okay thats what i got when i took the derivative of ((2t)^.5), which was 1/((2t)^.5). but what do you get when you multiply it by the derivative of arcsin, which is 1/((1-((2t)^2))^.5)
you should get,
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