The definition of the derivative of $f(x)$ is

$$f'(x) = \lim_{\Delta x \rightarrow 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}.$$

Use this definition to find the derivative of $f(x) = x - 5$. (You must use the definition.)

Answer:

YOu need to take the following limit

$$\lim_{h\rightarrow 0} \frac{f(x+h) - f(x)}{h} = \lim_{h\rightarrow 0} \frac{(x+h) - 5 - (x - 5)}{h} = \lim_{h\rightarrow 0} \frac{h}{h} = 1$$

Use the following rules of derivatives to find the 3 derivatives

$$\frac{d}{dx}[c f(x)] = c f'(x),\quad \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x),\quad \frac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x).$$

1. $f(x) = -16x^2 + 32 x - 12$, what is $f'(x)$?
   

   Answer: $-16(2 x ^1) + 32 (1) - 0 = -32x+32$

2. $f(x) = x^2 \sin(x) $, what is $f'(x)$?
   

   Answer: $(2x)\sin(x) + x^2 (\cos(x))$

3. $f(x) = x \sqrt{x}$, what is $f'(x)$?

   Answer: First, $f(x) = x^{3/2}$ by rewriting. So $f'(x) = (3/2)x^{3/2 - 1} = 3\sqsrt{x}/2$