## Calculus 8th Edition

$\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$
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: $\frac{2(sin(t) -sec^2(t))}{(cos(t)+tan(t))^3}$