How do you use the definition of the derivative to find #f'(x)# if #f(x)=1/sqrtx#?

1 Answer
Sep 28, 2016

Answer:

Start with #lim/(h→0){f(x+h)-f(x)}/h# multiply numerator and denominator by #f(x+h) + f(x)#

Explanation:

#lim/(h→0){(f(x+h))^2-(f(x))^2}/(h(f(x) + f(x+h)))#

We observe that #(f(x))^2 = 1/x# and #(f(x+h))^2 = 1/(x+h)# and we do the substitution:

#lim/(h→0){1/(x+h) - 1/x}/(h(1/sqrt(x) + 1/(sqrt(x+h)))#

Make a common denominator for #1/(x+h) - 1/x = (x)/((x(x +h))) - (x+h)/((x(x +h))) = -h/((x(x +h)))#

Substitute back into the limit:

#lim/(h→0){-h/((x(x +h)))}/(h(1/sqrt(x) + 1/(sqrt(x+h)))#

Please observe that #-h/h = -1#

#lim/(h→0){-1/((x(x +h)))}/((1/sqrt(x) + 1/(sqrt(x+h)))#

Now it is safe to let h go to zero:

#{-1/((x(x)))}/((1/sqrt(x) + 1/(sqrt(x)))) = (-1/x^2)/(2/sqrt(x)) = -1/(2x^(3/2))#