use the four - step definition of the derivative to find $f(x)$ if $f(x)=sqrt{7x}+1$. you may have to…

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 $h\rightarrow0$ of $\frac{f(x + h)-f(x)}{h}$: [ \begin{align*} f'(x)&=\lim_{h\rightarrow0}\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:

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