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

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Answer 1 · 2015-06-12T13:03:03

Rewrite the expression first.

$1 / (4-3t) + (3t )/ ((4-3t)^2) = (4-3t) / (4-3t)^2 + (3t )/ ((4-3t)^2) = 4/(4-3t)^2$

We can use the quotient rule at this point.

The derivative is:
$((0)(4-3t)^2-(4)[2(4-3t)*(-3)])/((4-3t)^2)^2 =(0+24(4-3t))/(4-3t)^4$

$= 24/(4-3t)^3$

Alternative

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

$4/(4-3t)^2 = 4(4-3t)^(-2)$

The derivative is may be found by using the power rule and the chain rule. $-8(4-3t)^(-3)*(-3) = 24(4-3t)^(-3)$

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