# How do you differentiate (1 / (4-3t)) + ( (3t )/ ((4-3t)^2))?

Jun 12, 2015

Rewrite the expression first.

#### Explanation:

$\frac{1}{4 - 3 t} + \frac{3 t}{{\left(4 - 3 t\right)}^{2}} = \frac{4 - 3 t}{4 - 3 t} ^ 2 + \frac{3 t}{{\left(4 - 3 t\right)}^{2}} = \frac{4}{4 - 3 t} ^ 2$

We can use the quotient rule at this point.

The derivative is:
$\frac{\left(0\right) {\left(4 - 3 t\right)}^{2} - \left(4\right) \left[2 \left(4 - 3 t\right) \cdot \left(- 3\right)\right]}{{\left(4 - 3 t\right)}^{2}} ^ 2 = \frac{0 + 24 \left(4 - 3 t\right)}{4 - 3 t} ^ 4$

$= \frac{24}{4 - 3 t} ^ 3$

Alternative

If (when) you've learned the chain rule, you may agree that it is easier to continue rewriting to get:

$\frac{4}{4 - 3 t} ^ 2 = 4 {\left(4 - 3 t\right)}^{- 2}$

The derivative is may be found by using the power rule and the chain rule. $- 8 {\left(4 - 3 t\right)}^{- 3} \cdot \left(- 3\right) = 24 {\left(4 - 3 t\right)}^{- 3}$