hi. i am unsure how to prove that f ' (x) = 1 / 2√(x - 1) is the derivative of f(x) = √(x - 1).
any help would be greatly appreciated. working out would be much appreciated.
cheers, jay.
Let f(x) = sqrt(x - 1).
Then the definition of the derivative says:
f'(x) = lim as h -> 0 of [f(x + h) - f(x)] / h.
Then:
[sqrt((x + h) - 1) - sqrt(x - 1)] / h.
Multiply both sides of the fraction by sqrt(x + h - 1) + sqrt(x - 1), giving us:
[sqrt(x + h - 1)^2 - sqrt(x - 1)^2] / [ sqrt(x + h - 1) + sqrt(x - 1) ], which comes out to:
[x + h - 1 - (x - 1)] / [h * sqrt(x + h - 1) + sqrt(x - 1)], which is:
h / [h * sqrt(x + h - 1) + sqrt(x - 1)], and we divide h out of both sides, giving us:
1 / [sqrt(x + h - 1) + sqrt(x - 1)]. As h gets very small, that is:
1 / 2(sqrt(x - 1)), and you're done.
Hope this helps.
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Verified answer
Let f(x) = sqrt(x - 1).
Then the definition of the derivative says:
f'(x) = lim as h -> 0 of [f(x + h) - f(x)] / h.
Then:
[sqrt((x + h) - 1) - sqrt(x - 1)] / h.
Multiply both sides of the fraction by sqrt(x + h - 1) + sqrt(x - 1), giving us:
[sqrt(x + h - 1)^2 - sqrt(x - 1)^2] / [ sqrt(x + h - 1) + sqrt(x - 1) ], which comes out to:
[x + h - 1 - (x - 1)] / [h * sqrt(x + h - 1) + sqrt(x - 1)], which is:
h / [h * sqrt(x + h - 1) + sqrt(x - 1)], and we divide h out of both sides, giving us:
1 / [sqrt(x + h - 1) + sqrt(x - 1)]. As h gets very small, that is:
1 / 2(sqrt(x - 1)), and you're done.
Hope this helps.