Using the limit definition, how do you differentiate #f(x)=sec x#?

1 Answer
Steve M
Mar 8, 2018

# f'(x) = d/dx sec x =tanx secx #

Explanation:

We have:

# f(x)=secx #

Using the limit definition of the derivative, we have:

# f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (sec(x+h)-sec(x))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (1/cos(x+h)-1/cos(x))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ((cosx-cos(x+h))/(cos(x+h)cos(x)))/h #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-cos(x+h))/(hcos(x+h)cos(x)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-(cosxcos h-sin x sin h))/(hcos(x+h)cos(x)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx-cosxcos h+sin x sin h)/(hcos(x+h)cos(x)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx(1-cos h)+sin x sin h)/(hcos(x+h)cos(x)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (cosx(1-cos h))/(hcos(x+h)cos(x))+(sin x sin h)/(hcos(x+h)cos(x)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) ((1-cos h))/(hcos(x+h))+(tan x sin h)/(hcos(x+h)) #

# \ \ \ \ \ \ \ \ = lim_(h rarr 0) (1-cos h)/h * sec(x+h)+(sin h)/h * tanx sec(x+h) #

Then we use two standard calculus limits:

# lim_(theta rarr 0) (1-cos theta)/theta = 0 # and # lim_(theta rarr 0) (sin theta)/theta = 1#

Which gives us:

# f'(x) = 0 * sec(x) + 1 * tanx sec(x) #
# \ \ \ \ \ \ \ \ = tanx secx #

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