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

1 Answer
Jan 12, 2017

#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))#