# How do you find the derivative of the function g(t)=(4t)/(t+1)?

Nov 9, 2016

$g ' \left(t\right) = \frac{4}{t + 1} ^ 2$

#### Explanation:

Use the quotient rule. The quotient rule states that the quotient of two functions, such as some function $y = \frac{u}{v}$, where $u$ and $v$ are functions of $x$, has a derivative of $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{du}}{\mathrm{dx}} \cdot v - u \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{v} ^ 2$.

So, applying this to the function at hand, we see that:

$g \left(t\right) = \frac{4 t}{t + 1}$

$g ' \left(t\right) = \frac{\left(\frac{d}{\mathrm{dt}} 4 t\right) \cdot \left(t + 1\right) - 4 t \left(\frac{d}{\mathrm{dt}} \left(t + 1\right)\right)}{t + 1} ^ 2$

$\textcolor{w h i t e}{g ' \left(t\right)} = \frac{4 \left(t + 1\right) - 4 t \left(1\right)}{t + 1} ^ 2$

$\textcolor{w h i t e}{g ' \left(t\right)} = \frac{4}{t + 1} ^ 2$