Derivative of a product involving e^x and sec(x)

Apply the product rule with $f(x) = e^x$ and $g(x) = \sec(x)$. We know $f'(x) = e^x$ and, using the memorized trigonometric derivative rule, $g'(x) = \sec(x)\tan(x)$.

Substitute into the product rule formula: $h'(x) = f'(x)g(x) + f(x)g'(x) = e^x \sec(x) + e^x \sec(x)\tan(x)$. This can optionally be factored as $h'(x) = e^x \sec(x)(1 + \tan(x))$.

The derivative of secant of x is secant of x times tangent of x, and this term is added, not subtracted. Unlike cosine, cosecant, and cotangent, secant's derivative carries a positive sign. This is why the final answer in part a has a plus sign between the two terms.