use the formula $f(x)=lim_{z\rightarrow x}\frac{f(z)-f(x)}{z - x}$ to find the derivative of the following…

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

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

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\rightarrow x}\frac{f(z)-f(x)}{z - x}\ &=\lim_{z\rightarrow x}\frac{(5+\sqrt{3z})-(5+\sqrt{3x})}{z - x}\ &=\lim_{z\rightarrow 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\rightarrow x}\frac{(\sqrt{3z}-\sqrt{3x})(\sqrt{3z}+\sqrt{3x})}{(z - x)(\sqrt{3z}+\sqrt{3x})}\ &=\lim_{z\rightarrow x}\frac{3z - 3x}{(z - x)(\sqrt{3z}+\sqrt{3x})}\ &=\lim_{z\rightarrow 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\neq x) when taking the limit) [ \begin{align*} &=\lim_{z\rightarrow 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}})