_Posted on September 28, 2024_ > [!theorem] **Theorem:** _Let $f$ be a real-valued function. If $f(x) = c$ for all $x$ in its domain, then $f$ is differentiable everywhere, and $f'(x) = 0$._ Proof: Given that $f(x) = c$, we can calculate $f'(x)$ as follows: $ f'(x) = \lim_{t \to x} \frac{f(t) - f(x)}{t - x} = \lim_{t \to x} \frac{c - c}{t - x} $ Since $c - c = 0$, the expression simplifies to: $ f'(x) = \lim_{t \to x} \frac{0}{t - x} = \lim_{t \to x} 0 \cdot \frac{1}{t - x} $ By [[Linearity of Limits]]: $ f'(x) = 0 \cdot \lim_{t \to x} \frac{1}{t - x} $ Finally, since multiplying by zero gives zero: $ f'(x) = 0 \qquad \blacksquare $ --- This can also be demonstrated as a special case of [[Derivative of Polynomials]] #TODO: