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

Question

Evelyn Agard · Accepted Answer

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)}#

Axel Alvarez · Answer

undefined To use the limit definition of the derivative to find the derivative of $f(x) = \sqrt{x - 4}$, we start by applying the definition:

$$
f'(x) = \lim_{{h \to 0}} \frac{{f(x + h) - f(x)}}{h}
$$

Substitute $f(x) = \sqrt{x - 4}$ into the formula:

$$
f'(x) = \lim_{{h \to 0}} \frac{{\sqrt{x + h - 4} - \sqrt{x - 4}}{h}}
$$

To simplify the expression, we multiply the numerator and denominator by the conjugate of the numerator:

$$
f'(x) = \lim_{{h \to 0}} \frac{{(\sqrt{x + h - 4} - \sqrt{x - 4})(\sqrt{x + h - 4} + \sqrt{x - 4})}}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}
$$

After expanding and simplifying, we get:

$$
f'(x) = \lim_{{h \to 0}} \frac{{(x + h - 4) - (x - 4)}}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}
$$

$$
f'(x) = \lim_{{h \to 0}} \frac{{x + h - 4 - x + 4}}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}
$$

$$
f'(x) = \lim_{{h \to 0}} \frac{h}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}
$$

$$
f'(x) = \lim_{{h \to 0}} \frac{1}{\sqrt{x + h - 4} + \sqrt{x - 4}}
$$

Now, we let $h$ approach 0:

$$
f'(x) = \frac{1}{\sqrt{x - 4} + \sqrt{x - 4}}
$$

$$
f'(x) = \frac{1}{2\sqrt{x - 4}}
$$

So, the derivative of $f(x) = \sqrt{x - 4}$ is $f'(x) = \frac{1}{2\sqrt{x - 4}}$.