Answer
$\dfrac{dy}{dx}=\dfrac{3}{2} [\tan(\sqrt{x})]^2 \sec^2(\sqrt{x})+\dfrac{\tan^3(\sqrt{x})}{2\sqrt{x}}$
Work Step by Step
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:
$\dfrac{dy}{dx}=\dfrac{3}{2} [\tan(\sqrt{x})]^2 \sec^2(\sqrt{x})+\dfrac{\tan^3(\sqrt{x})}{2\sqrt{x}}$
