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

Apr 24, 2016

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

Explanation:

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

So, using the power rule and chain rule:

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

Apr 24, 2016

$\frac{d y}{d x} = \frac{1}{x + 3} ^ 2$

Explanation:

$y = \frac{u}{v}$

$y ' = \frac{u ' \cdot v - v ' \cdot u}{{v}^{2}}$

$y = \frac{x + 2}{x + 3}$

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

$\frac{d y}{d x} = \frac{\left(x + 3\right) - \left(x + 2\right)}{x + 3} ^ 2$

$\frac{d y}{d x} = \frac{\cancel{x} + 3 - \cancel{x} - 2}{x + 3} ^ 2$

$\frac{d y}{d x} = \frac{1}{x + 3} ^ 2$