Show that the derivative of $y$ = tan $x$ is $$\boxed{ y^\prime =\dfrac{d}{dx}\tan x=\sec ^{2}x}$$

Solution $$\begin{eqnarray*} y^\prime &=&\dfrac{d}{dx}\tan x\underset{\underset{\color{#0066A7}{\hbox{Identity}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{d}{dx}\dfrac{\sin x}{\cos x}\underset{\underset{\color{#0066A7}{\hbox{Quotient Rule}}}{\color{#0066A7}{\uparrow }}}{=}\dfrac{\left[ \dfrac{d}{dx}\sin x\right] \cos x-\sin x\left[ \dfrac{d}{dx}\cos x\right] }{\cos ^{2}x} \\ &=&\dfrac{\cos x\cdot \cos x-\sin x\cdot (-\sin x)}{\cos ^{2}x}=\dfrac{\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}=\dfrac{1}{\cos ^{2}x}=\sec ^{2}x \end{eqnarray*}$$