Can anyone help me with this problem? I've been trying to figure it out but no dice.
Use the limit definition of the derivative to find f '(x) for f(x)= √(2x-1). An explanation would be greatly appreciated.
Use the limit definition of the derivative to find f '(x) for f(x)= √(2x-1). An explanation would be greatly appreciated.
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f '(x) = (f(x+h) - f(x))/h as the limit of h goes to 0.
This problem would start as [√(2(x+h)-1) - √(2x-1)]/h
From here you need to use the algebraic conjugate, √(2(x+h)-1) + √(2x-1) and multiply top and bottom by that. Get rid of the h by cancelling then sub in 0 for h when there isn't an h that causes problems, like if it makes the denominator 0.
This problem would start as [√(2(x+h)-1) - √(2x-1)]/h
From here you need to use the algebraic conjugate, √(2(x+h)-1) + √(2x-1) and multiply top and bottom by that. Get rid of the h by cancelling then sub in 0 for h when there isn't an h that causes problems, like if it makes the denominator 0.