# 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}##.

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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))’=-2cos(x)sin(x)##, we have

##frac{cos^3(x)+2sin^2(x)cos(x)}{cos^4(x)}##

Simplifying ##cos(x)##, we get

##frac{cos^2(x)+2sin^2(x)}{cos^3(x)}##

Writing ##cos^2(x)## as ##1-sin^2(x)##, we have

##frac{1-sin^2(x)+2sin^2(x)}{cos^3(x)}=frac{1+sin^2(x)}{cos^3(x)}##

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