# How do you use the limit definition to find the derivative of f(x)=(2-x)/(3x+1)?

Jan 21, 2017

$\frac{d}{\mathrm{dx}} \left(\frac{2 - x}{3 x + 1}\right) = - \frac{7}{3 x + 1} ^ 2$

#### Explanation:

By definition the derivative of $f \left(x\right)$ is:

${\lim}_{\Delta x \to 0} \frac{f \left(x + \Delta x\right) - f \left(x\right)}{\Delta x} = {\lim}_{\Delta x \to 0} \frac{\Delta f}{\Delta x}$

Calculate:

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

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

$\Delta f = \frac{\cancel{\left(2 - x\right) \left(3 x + 1\right)} - \Delta x \left(3 x + 1\right) - \cancel{\left(3 x + 1\right) \left(2 - x\right)} - 3 \Delta x \left(2 - x\right)}{\left(3 x + 3 \Delta x + 1\right) \left(3 x + 1\right)}$

$\Delta f = \frac{- \Delta x \left(3 x + 1 + 6 - 3 x\right)}{\left(3 x + 3 \Delta x + 1\right) \left(3 x + 1\right)} = - \frac{7 \Delta x}{\left(3 x + \Delta x + 1\right) \left(3 x + 1\right)}$

Divide by $\Delta x$:

$\frac{\Delta f}{\Delta x} = - \frac{7}{\left(3 x + \Delta x + 1\right) \left(3 x + 1\right)}$

Passing to the limit:

${\lim}_{\Delta x \to 0} \frac{\Delta f}{\Delta x} = {\lim}_{\Delta x \to 0} - \frac{7}{\left(3 x + \Delta x + 1\right) \left(3 x + 1\right)} = - \frac{7}{{\left(3 x + 1\right)}^{2}}$