3. (a) use the limit definition of the derivative to calculate the first derivative of $f(x)=\frac{2}{sqrt{3x…

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h\rightarrow0}\frac{f(x + h)-f(x)}{h}$.

Step2: Solve part (a)

Given $f(x)=\frac{2}{\sqrt{3x}}$, then $f(x + h)=\frac{2}{\sqrt{3(x + h)}}$. [ \begin{align*} f^\prime(x)&=\lim_{h\rightarrow0}\frac{\frac{2}{\sqrt{3(x + h)}}-\frac{2}{\sqrt{3x}}}{h}\ &=\lim_{h\rightarrow0}\frac{2\sqrt{3x}-2\sqrt{3(x + h)}}{h\sqrt{3x}\sqrt{3(x + h)}}\ &=\lim_{h\rightarrow0}\frac{2(\sqrt{3x}-\sqrt{3(x + h)})}{h\sqrt{3x}\sqrt{3(x + h)}}\times\frac{\sqrt{3x}+\sqrt{3(x + h)}}{\sqrt{3x}+\sqrt{3(x + h)}}\ &=\lim_{h\rightarrow0}\frac{2(3x-3(x + h))}{h\sqrt{3x}\sqrt{3(x + h)}(\sqrt{3x}+\sqrt{3(x + h)})}\ &=\lim_{h\rightarrow0}\frac{2(3x - 3x-3h)}{h\sqrt{3x}\sqrt{3(x + h)}(\sqrt{3x}+\sqrt{3(x + h)})}\ &=\lim_{h\rightarrow0}\frac{-6h}{h\sqrt{3x}\sqrt{3(x + h)}(\sqrt{3x}+\sqrt{3(x + h)})}\ &=\lim_{h\rightarrow0}\frac{-6}{\sqrt{3x}\sqrt{3(x + h)}(\sqrt{3x}+\sqrt{3(x + h)})}\ &=-\frac{1}{x\sqrt{3x}} \end{align*} ]

Step3: Solve part (b)

Given $f(x)=\sqrt{x}+x$, then $f(x + h)=\sqrt{x + h}+(x + h)$. [ \begin{align*} f^\prime(x)&=\lim_{h\rightarrow0}\frac{\sqrt{x + h}+(x + h)-(\sqrt{x}+x)}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{x + h}-\sqrt{x}+h}{h}\ &=\lim_{h\rightarrow0}\frac{\sqrt{x + h}-\sqrt{x}}{h}+1\ &=\lim_{h\rightarrow0}\frac{(\sqrt{x + h}-\sqrt{x})(\sqrt{x + h}+\sqrt{x})}{h(\sqrt{x + h}+\sqrt{x})}+1\ &=\lim_{h\rightarrow0}\frac{(x + h)-x}{h(\sqrt{x + h}+\sqrt{x})}+1\ &=\lim_{h\rightarrow0}\frac{h}{h(\sqrt{x + h}+\sqrt{x})}+1\ &=\lim_{h\rightarrow0}\frac{1}{\sqrt{x + h}+\sqrt{x}}+1\ &=\frac{1}{2\sqrt{x}}+1 \end{align*} ]

Answer:

(a) $f^\prime(x)=-\frac{1}{x\sqrt{3x}}$ (b) $f^\prime(x)=\frac{1}{2\sqrt{x}}+1$