Recall that \sec(x)=1/{\cos(x)}sec(x)=1cos(x). You can use the formula which states that
d/{dx} 1/{f(x)} = -\frac{f'}{f^2}.
This formula can easily be obtained by using the usual formula
d/{dx} {a(x)}/{b(x)}= \frac{a'\ b - a\ b'}{b^2}, where a(x)\equiv 1, and b(x)=f(x).
Since d/{dx} \cos(x)=-\sin(x), we have that
d/{dx} 1/{\cos(x)} = -\frac{-\sin(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)}
For the second derivative of \sec(x), let's derive one more time the first derivative: again, by the rule for the derivation of rational functions, we have
d/{dx} -\frac{\sin(x)}{\cos^2(x)} = \frac{\sin'(x)\cos^2(x) - \sin(x)(\cos^2(x))'}{\cos^4(x)}
Since \sin'(x)=\cos(x) and (\cos^2(x))'=-2\cos(x)\sin(x), we have
\frac{\cos^3(x)+2\sin^2(x)\cos(x)}{\cos^4(x)}
Simplifying \cos(x), we get
\frac{\cos^2(x)+2\sin^2(x)}{\cos^3(x)}
Writing \cos^2(x) as 1-\sin^2(x), we have
\frac{1-\sin^2(x)+2\sin^2(x)}{\cos^3(x)}=\frac{1+\sin^2(x)}{\cos^3(x)}
