Basic Differentiation Rules : Derivatives I don't know how to obtain the derivative of these equations.
Can you give me some hints, or some suggestions as to how to proceed?
$$(1)\;\;f(x) = \frac{x^2(2x+8)}{x+4}$$
$$ $$
$$(2)\;\;f(x) = (1/2+x) (1/3-x)$$
 A: $\require{cancel}$
$$f(x) = \dfrac{x^2(2x+8)}{x+4} = \dfrac{2x^2(\cancel{x + 4})}{\cancel{x+ 4}} = 2x^2, \;\;x \neq -4$$
I trust you can differentiate the right-hand side.
Note: In $(1)$, if there were no common factor to cancel, we could have used the quotient rule (and we could still use it, for practice!) If we call your function $f(x)$, and call the numerator $g(x)$ and the denominator $h(x)$, so that 
$$f(x) = \frac{g(x)}{h(x)}$$
then the quotient rule states that the derivative of $f(x) = g(x)/h(x)$ is given by
$$f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}.$$ 

In $(2)$ use the product rule, or else expand out the factors. The product rule tells us that if we have a function $f(x)$ that is the product of two smaller functions $g(x), \, h(x)$, then 
$$f'(x) = (g\cdot h)'=g'\cdot h+g\cdot h'$$
Note that in our case, $f(x) = (1/2+x) (1/3-x)$, so put $g(x) = \frac 12 + x$ and $h(x) = \frac 13 - x$. Then $$\begin{align} f'(x) & = g'(x) \cdot h(x) + g(x) \cdot h'(x) \\ \\ 
& = 1\cdot \left(\frac 13 - x\right) + \left(\frac 12 + x\right) \cdot (-1) \\ \\
& = \frac 13 - x - \frac 12 - x \\ \\ & = -\frac 16 - 2x\end{align}$$
If we expand the factors, $$f(x) = (1/2+x) (1/3-x) = \frac 16 + -\frac 16 x - x^2$$ Again, the derivative of $f(x)$ should be fairly straightforward, and you can double check our earlier work using the product rule to confirm the results are the same.
A: Hint:
$$ (\frac{u}{v})' = \frac{u'v - uv'}{v^2} $$
$$ (uv)' = u'v + uv' $$
