1. The problem: find the derivative of $\tan x$ with respect to $x$.\n\n2. Recall the definition: $\tan x = \frac{\sin x}{\cos x}$.\n\n3. Use the quotient rule: if $f(x) = \frac{g(x)}{h(x)}$, then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{h(x)^2}$.\n\n4. Let $g(x) = \sin x$ and $h(x) = \cos x$. Then $g'(x) = \cos x$ and $h'(x) = -\sin x$.\n\n5. Substitute into the quotient rule:\n$$\frac{d}{dx} \tan x = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}.$$\n\n6. Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$, so numerator becomes 1.\n\n7. Simplify the expression: $$\frac{1}{\cos^2 x} = \sec^2 x.$$\n\n8. Final answer: $$\frac{d}{dx} \tan x = \sec^2 x.$$