The idea: To prove that some expression involving f, f', and x equals zero, construct a function phi(x) whose derivative is that expression (up to a nonzero factor), then apply Rolle's to phi.

Pattern 1: f'(c) + kf(c) = 0 Use phi(x) = e^(kx) * f(x). Then phi'(x) = e^(kx)(f'(x) + kf(x)). If phi(a) = phi(b), Rolle's gives phi'(c) = 0, so f'(c) + kf(c) = 0.

Pattern 2: f'(c) = cf(c) Use phi(x) = f$\frac{x}{e}$^($x^2$/2). Then phi'(x) = (f'(x) - xf(x))/e^($x^2$/2).

Pattern 3: nf(c) + cf'(c) = 0 Use phi(x) = $x^n$ * f(x). Then phi'(x) = x^(n-1)(nf(x) + xf'(x)).

How to find the right auxiliary function: Look at the target equation and try to recognize it as the derivative of a product, quotient, or composition. If the equation is f'/f = g'/g, then d/dx[ln$\frac{f}{g}$] = 0, so use phi = $\frac{f}{g}$.