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What is the second derivative of #secx#? 

Answer 1

Emily Ammerman

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)}#

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Answer 2

Luke Alvarez

The second derivative of sec(x) is sec(x) * tan(x)^2 + sec(x) * sec(x).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

What is the second derivative of #secx#?&nbsp;

What is the second derivative of #secx#?