How do you use the limit definition of the derivative to find the derivative of $f(x)=1/sqrtx$ ?

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Answer 1 · 2017-04-25T23:19:04

$f'(x) = -1/(2xsqrtx)$

By definition of derivative we have:

$f'(x) = lim_(h->0) (f(x+h)-f(x))/h$

So for $f(x) =1/sqrtx$ :

$f'(x) = lim_(h->0) (1/sqrt(x+h)-1/sqrt(x))/h$

$f'(x) = lim_(h->0) (sqrtx-sqrt(x+h))/(hsqrtxsqrt(x+h))$

Rationalize the numerator using the identity: $(a+b)(a-b) = a^2-b^2$ :

$f'(x) = lim_(h->0) (sqrtx-sqrt(x+h))/(hsqrtxsqrt(x+h)) xx (sqrtx+sqrt(x+h))/(sqrtx+sqrt(x+h))$

$f'(x) = lim_(h->0) (x-x-h)/((hsqrtxsqrt(x+h))(sqrtx+sqrt(x+h)))$

$f'(x) = lim_(h->0) -cancelh/((cancelhsqrtxsqrt(x+h))(sqrtx+sqrt(x+h)))$

$f'(x) = -1/((sqrtxsqrt(x))(sqrtx+sqrt(x)))$

$f'(x) = -1/(2xsqrtx)$

How do you use the limit definition of the derivative to find the derivative of f(x)=1/sqrtxf(x)=1/sqrtx?