Question

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

Answer 1

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

Explanation:

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)`