Chapter 3: Derivatives - Section 3.5 - Derivatives of Trigonometric Functions - Exercises 3.5 - Page 143: 67

See graph and explanations.

See graph; the minimum of the derivative can be found at $x=0$, which gives $y'=sec^2(0)=1$, and it does not appear to have a maximum value because $y'=sec^2(x)$ can be as large as possible in this region. The slope will never be negative because the derivative is a square function, that is $y'=sec^2x\geq1$