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derivatives of secx by first principle

derivatives of secx by first principle


1 Answers

Aman Bansal
592 Points
8 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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Aman Bansal

Askiitian Expert

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