## Calculus, 10th Edition (Anton)

$\dfrac{dy}{dx}=\dfrac{3}{2} [\tan(\sqrt{x})]^2 \sec^2(\sqrt{x})+\dfrac{\tan^3(\sqrt{x})}{2\sqrt{x}}$
In order to derivate this function, you have to use product rule: $\dfrac{d}{dx}(a*b)=a'b+ab'$ So let's identify a and b and derivate them $a = \sqrt{x}$ $a' = \dfrac{1}{2\sqrt{x}}$ $b =(\tan(\sqrt{x}))^3$ $b'= \dfrac{3[\tan(\sqrt{x})]^2\sec^2(\sqrt{x})}{2\sqrt{x}}$ *Note: Here you have to apply chain rule twice in order to derivate B Substitute in the formula: $\dfrac{dy}{dx} =(\dfrac{1}{2\sqrt{x}})(\tan(\sqrt{x}))^3+ (\sqrt{x})(\dfrac{3[\tan(\sqrt{x})]^2\sec^2(\sqrt{x})}{2\sqrt{x}})$ Simplify and get the answer: $\dfrac{dy}{dx}=\dfrac{3}{2} [\tan(\sqrt{x})]^2 \sec^2(\sqrt{x})+\dfrac{\tan^3(\sqrt{x})}{2\sqrt{x}}$