Question

use the four - step definition of the derivative to find $f(x)$ if $f(x)=sqrt{7x}+1$. you may have to simplify your answers to recieve credit.
$f(x + h)=
$f(x + h)-f(x)=
$\frac{f(x + h)-f(x)}{h}=
$f(x)=$

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$:
$f(x + h)=\sqrt{7(x + h)}+1$

Step2: Calculate $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=\sqrt{7(x + h)}+1-(\sqrt{7x}+1)\\ &=\sqrt{7(x + h)}-\sqrt{7x} \end{align*}$$

\]

Step3: Compute $\frac{f(x + h)-f(x)}{h}$

\[

$$\begin{align*} \frac{f(x + h)-f(x)}{h}&=\frac{\sqrt{7(x + h)}-\sqrt{7x}}{h}\\ &=\frac{(\sqrt{7(x + h)}-\sqrt{7x})(\sqrt{7(x + h)}+\sqrt{7x})}{h(\sqrt{7(x + h)}+\sqrt{7x})}\\ &=\frac{7(x + h)-7x}{h(\sqrt{7(x + h)}+\sqrt{7x})}\\ &=\frac{7x+7h - 7x}{h(\sqrt{7(x + h)}+\sqrt{7x})}\\ &=\frac{7h}{h(\sqrt{7(x + h)}+\sqrt{7x})}\\ &=\frac{7}{\sqrt{7(x + h)}+\sqrt{7x}} \end{align*}$$

\]

Step4: Find $f'(x)$

Take the limit as $h
ightarrow0$ of $\frac{f(x + h)-f(x)}{h}$:
\[

$$\begin{align*} f'(x)&=\lim_{h ightarrow0}\frac{7}{\sqrt{7(x + h)}+\sqrt{7x}}\\ &=\frac{7}{\sqrt{7x}+\sqrt{7x}}\\ &=\frac{7}{2\sqrt{7x}}\\ &=\frac{\sqrt{7}}{2\sqrt{x}} \end{align*}$$

\]

Answer:

$f(x + h)=\sqrt{7(x + h)}+1$
$f(x + h)-f(x)=\sqrt{7(x + h)}-\sqrt{7x}$
$\frac{f(x + h)-f(x)}{h}=\frac{7}{\sqrt{7(x + h)}+\sqrt{7x}}$
$f'(x)=\frac{\sqrt{7}}{2\sqrt{x}}$