## Calculus: Early Transcendentals 8th Edition

$y'=-\dfrac{3(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})}{2\sqrt{\tan3x}}$
$y=\cos(e^{\sqrt{\tan3x}})$ Start the differentiation process by using the chain rule: $y'=-\sin(e^{\sqrt{\tan3x}})(e^{\sqrt{\tan3x}})'=...$ Apply the chain rule one more time to evaluate the indicated derivative: $...=-(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})(\sqrt{\tan3x})'=...$ $...=-(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})[(\tan3x)^{1/2}]'=...$ Once again, apply the chain rule to evaluate the indicated derivative: $...=-\dfrac{1}{2}(\tan3x)^{-1/2}(\tan3x)'(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})=...$ Apply the chain rule one last time to evaluate the indicated derivative and simplify the expression: $...=-\dfrac{1}{2}(3x)'(\tan3x)^{-1/2}(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})=...$ $...=-\dfrac{3(\sec^{2}3x)(e^{\sqrt{\tan3x}})\sin(e^{\sqrt{\tan3x}})}{2\sqrt{\tan3x}}$