# How do you calculate the derivate for f(x)=[(x+2)/(x+1)](2x-7)?

Aug 1, 2015

Using a combination of Product Rule and Quotient Rule.

#### Explanation:

Product Rule:
If $f \left(x\right) = g \left(x\right) \setminus \cdot h \left(x\right)$,
then ${f}^{\setminus} p r i m e \left(x\right) = g \left(x\right) \setminus \cdot h \setminus p r i m e \left(x\right) + h \left(x\right) \setminus \cdot g \setminus p r i m e \left(x\right)$

Quotient Rule:
If $f \left(x\right) = \setminus \frac{g \left(x\right)}{h \left(x\right)}$
then ${f}^{\setminus} p r i m e \left(x\right) = \setminus \frac{h \left(x\right) \setminus \cdot g \setminus p r i m e \left(x\right) - g \left(x\right) \setminus \cdot h \setminus p r i m e \left(x\right)}{h {\left(x\right)}^{2}}$

Now, you can express $f \left(x\right)$ as

$f \left(x\right) = \setminus {\underbrace{g \left(x\right) \setminus \cdot h \left(x\right)}}_{\setminus} \textrm{\prod u c t r \underline{e}}$,

where $g \left(x\right) = \setminus {\underbrace{\setminus \frac{x + 2}{x + 1}}}_{\setminus} \textrm{q u o t i e n t r \underline{e}}$ and $h \left(x\right) = 2 x - 7$

OR as

$f \left(x\right) = \setminus {\underbrace{\setminus \frac{g \left(x\right)}{h \left(x\right)}}}_{\setminus} \textrm{q u o t i e n t r \underline{e}}$,

where $g \left(x\right) = \setminus {\underbrace{\left(x + 2\right) \left(2 x - 7\right)}}_{\setminus} \textrm{\prod u c t r \underline{e}}$ and $h \left(x\right) = x + 1$.

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