1. **State the problem:** Differentiate the function $f(x) = \tan x$ with respect to $x$. 2. **Recall the formula:** The derivative of $\tan x$ is given by the formula $$\frac{d}{dx}(\tan x) = \sec^2 x$$ This comes from the quotient rule since $\tan x = \frac{\sin x}{\cos x}$. 3. **Explain the differentiation:** Using the quotient rule, $$\frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$$ 4. **Simplify using the Pythagorean identity:** Since $\sin^2 x + \cos^2 x = 1$, we have $$\frac{1}{\cos^2 x} = \sec^2 x$$ 5. **Final answer:** Therefore, $$\frac{d}{dx}(\tan x) = \sec^2 x$$ This means the slope of the tangent line to the curve $y=\tan x$ at any point $x$ is $\sec^2 x$.