# How do you find the derivative of (x-3)/(2x+1)?

Oct 5, 2017

$\frac{7}{2 x + 1} ^ 2$

#### Explanation:

$\text{differentiate using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = x - 3 \Rightarrow g ' \left(x\right) = 1$

$h \left(x\right) = 2 x + 1 \Rightarrow h ' \left(x\right) = 2$

$\Rightarrow f ' \left(x\right) = \frac{\left(2 x + 1\right) - 2 \left(x - 3\right)}{2 x + 1} ^ 2$

$\textcolor{w h i t e}{\Rightarrow f ' \left(x\right)} = \frac{7}{2 x + 1} ^ 2$

Oct 5, 2017

$= \frac{7}{4 {x}^{2} - 4 x + 1}$ or $\frac{7}{2 x + 1} ^ 2$

#### Explanation:

Quotient rule: $\left(u ' v\right) - \left(u v '\right)$ all divided by ${v}^{2}$, assuming the top line is $u$ and bottom is $v$

In this function, $u = x - 3$ and $v = 2 x + 1$

$\therefore u ' = 1 \text{ and } v ' = 2$

So plugging that in gets:

$\frac{1 \cdot \left(2 x + 1\right) - \left(x - 3\right) \cdot 2}{2 x + 1} ^ 2$

$= \frac{\left(2 x + 1\right) - \left(2 x - 6\right)}{4 {x}^{2} + 4 x + 1}$

$= \frac{\cancel{2 x} + 1 - \cancel{2 x} + 6}{4 {x}^{2} - 4 x + 1}$

$= \frac{7}{4 {x}^{2} - 4 x + 1}$