Question

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

(b) use the limit definition of the derivative to calculate $\frac{df}{dx}$ where $f(x)=sqrt{x}+x$.

Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative of a function $y = f(x)$ is $f^\prime(x)=\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}$.

Step2: Solve part (a)

Given $f(x)=\frac{2}{\sqrt{3x}}$, then $f(x + h)=\frac{2}{\sqrt{3(x + h)}}$.

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Step3: Solve part (b)

Given $f(x)=\sqrt{x}+x$, then $f(x + h)=\sqrt{x + h}+(x + h)$.

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Answer:

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