use the formal definition to find the derivative of ( g(x)=sqrt{6x} ) at ( x = 4 ).\nfind ( g^{prime}(x) )…

Explanation:

Step1: Recall the formal definition of the derivative

The formal definition of the derivative is (g^{\prime}(x)=\lim_{h\rightarrow0}\frac{g(x + h)-g(x)}{h}). Given (g(x)=\sqrt{6x}), then (g(x + h)=\sqrt{6(x + h)}). So, (\frac{g(x + h)-g(x)}{h}=\frac{\sqrt{6(x + h)}-\sqrt{6x}}{h}).

Step2: Rationalize the numerator

Multiply the numerator and denominator by (\sqrt{6(x + h)}+\sqrt{6x}). [ \begin{align*} \frac{\sqrt{6(x + h)}-\sqrt{6x}}{h}\times\frac{\sqrt{6(x + h)}+\sqrt{6x}}{\sqrt{6(x + h)}+\sqrt{6x}}&=\frac{6(x + h)-6x}{h(\sqrt{6(x + h)}+\sqrt{6x})}\ &=\frac{6x+6h - 6x}{h(\sqrt{6(x + h)}+\sqrt{6x})}\ &=\frac{6h}{h(\sqrt{6(x + h)}+\sqrt{6x})} \end{align*} ] Cancel out the (h) (since (h\neq0) when taking the limit), we get (\frac{6}{\sqrt{6(x + h)}+\sqrt{6x}}).

Step3: Take the limit as (h\rightarrow0)

(\lim_{h\rightarrow0}\frac{6}{\sqrt{6(x + h)}+\sqrt{6x}}=\frac{6}{\sqrt{6x}+\sqrt{6x}}=\frac{6}{2\sqrt{6x}}=\frac{3}{\sqrt{6x}}=\frac{\sqrt{6}}{2\sqrt{x}}) (rationalizing (\frac{3}{\sqrt{6x}}=\frac{3\sqrt{6}}{6\sqrt{x}}=\frac{\sqrt{6}}{2\sqrt{x}})).

Answer:

(g^{\prime}(x)=\frac{\sqrt{6}}{2\sqrt{x}})