Question

question
find the derivative of the function (f(x)=sqrt{x - 6}) using the limit definition of the derivative.
provide your answer below:
f(x)=□

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^{\prime}(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$. Given $f(x)=\sqrt{x - 6}$, then $f(x + h)=\sqrt{(x + h)-6}=\sqrt{x+h - 6}$.

Step2: Substitute into the formula

$\frac{f(x + h)-f(x)}{h}=\frac{\sqrt{x + h-6}-\sqrt{x - 6}}{h}$.

Step3: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator $\sqrt{x + h-6}+\sqrt{x - 6}$. We get $\frac{(\sqrt{x + h-6}-\sqrt{x - 6})(\sqrt{x + h-6}+\sqrt{x - 6})}{h(\sqrt{x + h-6}+\sqrt{x - 6})}$.
Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(x + h-6)-(x - 6)=x + h-6-x + 6=h$.
So the expression is $\frac{h}{h(\sqrt{x + h-6}+\sqrt{x - 6})}=\frac{1}{\sqrt{x + h-6}+\sqrt{x - 6}}$.

Step4: Find the limit as $h

ightarrow0$
$f^{\prime}(x)=\lim_{h
ightarrow0}\frac{1}{\sqrt{x + h-6}+\sqrt{x - 6}}$. As $h
ightarrow0$, we substitute $h = 0$ into the expression, and we get $f^{\prime}(x)=\frac{1}{2\sqrt{x - 6}}$.

Answer:

$\frac{1}{2\sqrt{x - 6}}$