Question

How do you find f'(x) using the limit definition given `(2/sqrt x)`?

Answer 1

`f'(x) = -1/(xsqrt(x))`

Explanation:

The limit definition of the derivative of a function `f(x)` is:

`f'(x) = lim_(Deltax->0) (f(x+Deltax)-f(x))/(Deltax) = lim_(Deltax->0) (Deltaf)/(Deltax)`

Let's calculate the function variation between `x` and `x+Deltax`:

`Delta f = 2/sqrt(x+Deltax) -2/sqrt(x)= 2(1/sqrt(x+Deltax) -1/sqrt(x))= 2(sqrt(x)- sqrt(x+Deltax))/(sqrt(x)sqrt(x+Deltax))`

We can now rationalize the numerator:

`Delta f = 2(sqrt(x)- sqrt(x+Deltax))/(sqrt(x)sqrt(x+Deltax)) * (sqrt(x)+ sqrt(x+Deltax))/(sqrt(x)+ sqrt(x+Deltax))= 2(x-(x+Deltax))/(xsqrt(x+Deltax) + sqrt(x) (x+Deltax))= (-2Deltax)/(xsqrt(x+Deltax) + sqrt(x) (x+Deltax))`

Divide by `Deltax`:

`(Deltaf)/(Deltax) = (-2)/(xsqrt(x+Deltax) + sqrt(x) (x+Deltax))`

and passing to the limit for `Deltax->0`:

`lim_(Deltax->0) (-2)/(xsqrt(x+Deltax) + sqrt(x) (x+Deltax)) = (-2)/(xsqrt(x) +sqrt(x)x) = (-2)/(2xsqrt(x)) = -1/(xsqrt(x))`