Using the limit definition, how do you differentiate #f(x) =1/sqrt(x) #?

1 Answer
Truong-Son N.
Dec 7, 2015

The limit definition is:

#lim_(h->0) (f(x+h) - f(x))/(h)#

If you think of it in the following way, it should help:

  1. Take a regular graph and choose any point on it.
  2. Zoom into it. Either use your calculator or just imagine it.

You've just simulated the limit definition of the derivative. The more you zoom in, the more linear the graph looks. Then it's just the slope equation after you've zoom in a lot. That's the idea.

Really, the main trick to doing problems like this is being able to work with cross-multiplying and complex conjugates.

#f'(x)#

#= lim_(h->0) (1/(sqrt(x+h)) - 1/sqrt(x))/(h)#

Move the #h# into the numerator functions' denominators.

#= lim_(h->0) 1/(hsqrt(x+h)) - 1/(hsqrt(x))#

Get common denominators by cross-multiplying.

#= lim_(h->0) 1/(hsqrt(x+h))*(sqrtx)/(sqrtx) - 1/(hsqrt(x))*sqrt(x+h)/sqrt(x+h)#

#= lim_(h->0) sqrt(x)/(hsqrt(x+h)sqrt(x)) - sqrt(x+h)/(hsqrt(x+h)sqrt(x))#

Multiply by the complex conjugate of the resultant numerator as a fraction equal to #1#.

#= lim_(h->0) (sqrt(x) - sqrt(x+h))/(hsqrt(x(x+h))) * (sqrt(x) + sqrt(x+h))/(sqrt(x) + sqrt(x+h))#

#= lim_(h->0) (x - (x+h))/((hsqrt(x(x+h)))(sqrt(x) + sqrt(x+h))#

Thus, you get a cancellation of #x# in the numerator and #h# is by itself. Now you can cancel #h# and finish the evaluation without getting an undefined answer.

#= lim_(h->0) (-cancel(h))/((cancel(h)sqrt(x(x+h)))(sqrt(x) + sqrt(x+h))#

#= lim_(h->0) (-1)/(sqrt(x(x+cancel(h)^(0)))(sqrt(x) + sqrt(x+cancel(h)^(0)))#

#= (-1)/(2xsqrt(x))#

#= color(blue)(-(1)/(2x^"3/2"))#

Naturally, if you check using the power rule and using #1/(sqrtx) = x^"-1/2"#, you will get the same answer.

#f'(x) = -1/2 * x^("-1/2" - 1) = color(blue)(-1/(2x^"3/2"))#

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