How do you use the limit definition of the derivative to find the derivative of #f(x)=sqrt(x-4)#?

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Answer 1 · 2016-11-02T08:28:23

Please see the explanation for details on how you do the requested process.

The limit definition is:

#lim_(hto0){f(x+h) - f(x)}/h#

Given:

#f(x) = sqrt(x - 4)#

To obtain #f(x + h), #substitute #x + h# for #x#:

#f(x + h) = sqrt(x + h - 4)#

Substituting into the definition:

#lim_(hto0){sqrt(x + h - 4) - sqrt(x - 4)}/h#

Because we know that #(a - b)(a + b) = a^2 - b^2#, we choose to multiply by #{sqrt(x + h - 4) + sqrt(x - 4)}/{sqrt(x + h - 4) + sqrt(x - 4)}#:

#lim_(hto0){sqrt(x + h - 4) - sqrt(x - 4)}/h{sqrt(x + h - 4) + sqrt(x - 4)}/{sqrt(x + h - 4) + sqrt(x - 4)}#

This will square the square roots in the numerator and, eventually, leave nothing but h:

#lim_(hto0){(sqrt(x + h - 4))^2 - (sqrt(x - 4))^2}/(h{sqrt(x + h - 4) + sqrt(x - 4)})#

Squaring the square roots makes them disappear:

#lim_(hto0){x + h - 4 - (x - 4)}/(h{sqrt(x + h - 4) + sqrt(x - 4)})#

Distribute the - through the ()s in the numerator:

#lim_(hto0){x + h - 4 - x + 4}/(h{sqrt(x + h - 4) + sqrt(x - 4)})#

The numerator simplifies to become only h:

#lim_(hto0)h/(h{sqrt(x + h - 4) + sqrt(x - 4)})#

#h/h# becomes 1:

#lim_(hto0)1/{sqrt(x + h - 4) + sqrt(x - 4)}#

Now, it is ok to let #hto0#:

#1/{sqrt(x - 4) + sqrt(x - 4)}#

Combine the terms in the denominator:

#1/{2sqrt(x - 4)}#