Answer
$\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$
Work Step by Step
Original Expression: $A(t) = \frac{1}{(cos(t) + tan(t))^2}$
$u = cos(t) + tan(t)$
$A(u) = \frac{1}{x^2}$ => $x^{-2}$
Apply the chain rule: A’(u)u’
$A’(u) = -2(u)^{-3}(-sin(t) + sec^2(t))$
$A’(t) = -2(cos(t) + tan(t))^{-3}(-sin(t) + sec^2(t))$
Simplify:
$\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$
