# derivatives of secx by first principle

Aman Bansal
592 Points
10 years ago

Dear Akash,

Since you want to differentiate using first principles, differentiate using the definition of the derivative.

If f(x) = sec(x), then:
f''(x) = lim (h-->0) [sec(x + h) - sec(x)]/h.

Then, since sec(x + h) = 1/cos(x + h) and sec(x) = 1/cos(x):
f''(x) = lim (h-->0) [sec(x + h) - sec(x)]/h
= lim (h-->0) [1/cos(x + h) - 1/cos(x)]/h
= lim (h-->0) [cos(x) - cos(x + h)]/[h*cos(x)cos(x + h)].
(Note that I multiplied the numerator and denominator by cos(x)cos(x + h).)

By the cosine addition formula, we can write:
f''(x) = lim (h-->0) [cos(x) - cos(x + h)]/[h*cos(x)cos(x + h)]
= lim (h-->0) {cos(x) - [cos(x)cos(h) - sin(x)sin(h)]}/[h*cos(x)cos(x + h)]
= lim (h-->0) [cos(x) - cos(x)cos(h) + sin(x)sin(h)]/[h*cos(x)cos(x + h)]
= lim (h-->0) {cos(x)[1 - cos(h)] + sin(x)sin(h)}/[h*cos(x)cos(x + h)], by factoring out cos(x)
= lim (h-->0) {cos(x)[1 - cos(h)]/h + sin(x)[sin(h)/h]}/[cos(x)cos(x + h)].

Then, since lim (x-->0) [1 - cos(x)]/x = 0 and lim (x-->0) sin(x)/x = 1.
f''(x) = lim (h-->0) {cos(x)[1 - cos(h)]/h + sin(x)[sin(h)/h]}/[cos(x)cos(x + h)]
= [cos(x)(0) + sin(x)(1)]/cos^2(x)
= sin(x)/cos^2(x)
= sec(x)tan(x).

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Thanks

Aman Bansal