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find derivative using limit definition: f(x) = x - sqrt(x) so f'(x) = lim h->0 [f(x+h) - f(x)]/h but I keep trying to solve by...Asked by terra
find derivative using limit definition:
f(x) = x - sqrt(x)
so f'(x) =
lim
h->0 [f(x+h) - f(x)]/h
but I keep trying to solve by multiplying by the conjugate but I can't figure it out..there's nothing that can be cancelled or anything and I can't get the derivative
sorry this is a repost, but i messed up my other one and i really need help on this
f(x) = x - sqrt(x)
so f'(x) =
lim
h->0 [f(x+h) - f(x)]/h
but I keep trying to solve by multiplying by the conjugate but I can't figure it out..there's nothing that can be cancelled or anything and I can't get the derivative
sorry this is a repost, but i messed up my other one and i really need help on this
Answers
Answered by
Steve
The key to this is the binomial theorem:
(x+h)^n = x^n + n*x^(n-1)*h + n(n-1)/2 * x^(n-2) * h^2 + ...
So, we find ourselves with
(h + sqrt(x) - sqrt(x+h))/h
= (h + x^(1/2) - (x^(1/2) + 1/2 * x^(-1/2)*h + <higher powers of h>)/h
= 1 - 1/2 * x^(-1/2)
(all terms with h go to zero)
(x+h)^n = x^n + n*x^(n-1)*h + n(n-1)/2 * x^(n-2) * h^2 + ...
So, we find ourselves with
(h + sqrt(x) - sqrt(x+h))/h
= (h + x^(1/2) - (x^(1/2) + 1/2 * x^(-1/2)*h + <higher powers of h>)/h
= 1 - 1/2 * x^(-1/2)
(all terms with h go to zero)
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