What is the second derivative of ##secx##? 

Recall that ##\sec(x)=1/{\cos(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)}##