SOLUTION: find the derivative of the following function:
y=(2x+1/2x-1)^1/2
y=sin3t/(t+1)^1/2
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Question 1122716: find the derivative of the following function:
y=(2x+1/2x-1)^1/2
y=sin3t/(t+1)^1/2
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
It is (1/2)[(2x+1)/(2x-1)]^(-1/2) * derivative of (2x+1)/(2x-1)
That derivative is {1/(2x-1)^2}}*(2x-1)*2-(2x+1)(2)=-4/(2x-1)^2
={(-2)/2x-1)^2}*1/sqrt((2x+1)/(2x-1))
The -4 becomes -2 because of the (1/2) in the first part of the derivative.
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