use the formula $f(x)=lim_{z
ightarrow x}\frac{f(z)-f(x)}{z - x}$ to find the derivative of the following function.
$f(x)=5+sqrt{3x}$
$f(x)=square$ (type an exact answer, using radicals as needed.)

Question

Accepted Answer

# Explanation:
## Step1: Find f(z)
Given $f(x)=5 + \sqrt{3x}$, then $f(z)=5+\sqrt{3z}$.
## Step2: Substitute into the derivative formula
$$
\begin{align*}
f^{\prime}(x)&=\lim_{z
ightarrow x}\frac{f(z)-f(x)}{z - x}\
&=\lim_{z
ightarrow x}\frac{(5+\sqrt{3z})-(5+\sqrt{3x})}{z - x}\
&=\lim_{z
ightarrow x}\frac{\sqrt{3z}-\sqrt{3x}}{z - x}
\end{align*}
$$
## Step3: Rationalize the numerator
Multiply the numerator and denominator by $\sqrt{3z}+\sqrt{3x}$
$$
\begin{align*}
&=\lim_{z
ightarrow x}\frac{(\sqrt{3z}-\sqrt{3x})(\sqrt{3z}+\sqrt{3x})}{(z - x)(\sqrt{3z}+\sqrt{3x})}\
&=\lim_{z
ightarrow x}\frac{3z - 3x}{(z - x)(\sqrt{3z}+\sqrt{3x})}\
&=\lim_{z
ightarrow x}\frac{3(z - x)}{(z - x)(\sqrt{3z}+\sqrt{3x})}
\end{align*}
$$
## Step4: Simplify and find the limit
Cancel out $z - x$ (since $z
eq x$ when taking the limit)
$$
\begin{align*}
&=\lim_{z
ightarrow x}\frac{3}{\sqrt{3z}+\sqrt{3x}}\
&=\frac{3}{\sqrt{3x}+\sqrt{3x}}\
&=\frac{3}{2\sqrt{3x}}
\end{align*}
$$

# Answer:
$\frac{3}{2\sqrt{3x}}$

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QUESTION IMAGE

Question

Explanation:

Step1: Find f(z)

Step2: Substitute into the derivative formula

Step3: Rationalize the numerator

Step4: Simplify and find the limit

Answer: